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Regular practice with AP Inter 2nd Year Physics Study Material Chapter 6 Electromagnetic Induction Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Physics 6th Lesson Electromagnetic Induction Questions and Answers

I. Multiple Choice Questions

Question 1.
Faraday’s law of electromagnetic induction states that the induced emf is proportional to:
1) the rate of change of magnetic flux
2) the absolute value of the magnetic field
3) the distance between coils
4) the square of the current
Answer:
1) the rate of change of magnetic flux
From Faraday’s second law emf e = -N \(\frac{d \phi}{d t}\) ⇒ e ∝ \(\frac{d \phi}{d t}\) = Rate of change in flux.

Question 2.
Lenz’s law is a consequence of which principle?
1) Conservation of energy
2) Newton’s first law
3) Coulomb’s law
4) Ohm’s law
Answer:
1) Conservation of energy
Lenz’s Law is a consequence of Law of conservation of energy.

Question 3.
A conducting loop is rotated in a uniform magnetic field. The induced emf will be maximum when the plane of the loop is:
1) parallel to the magnetic field
2) perpendicular to the magnetic field
3) at 45° to the magnetic field
4) at any angle
Answer:
1) parallel to the magnetic field
Induced emf in a rotating loop is e = NB Asinθ where θ is the angle between the normal to the loop and the magnetic field. The emf is maximum when sinθ = 1 ⇒ θ = 90°. That means the normal to the loop is perpendicular to the field, so the plane of the loop is parallel to the magnetic field.

Question 4.
The SI unit of self inductance is:
1) Tesla
2) Weber
3) Ampere
4) Henry
Answer:
4) Henry
Units of Self Induction : Henry.

Question 5.
A coil of 500 turns and area of 0.01 m² is placed perpendicular to magnetic field. If the magnetic field changes uniformly from 0 to 0.5 T in 0.1 s, what is the average emf induced?
1) 0.25 V
2) 2.5 V
3) 25 V
4) 50 V
Answer:
3) 25 V
N = 500, A = 0.01m² = 1 × 10^-2m², \(\frac{\mathrm{dB}}{\mathrm{dt}}=\frac{0.5}{0.1}\) = 5 Emf e = NA\(\frac{\mathrm{dB}}{\mathrm{dt}}\) = 500 × 10^-2 × 5 = 25N

Question 6.
Mutual induction between two coils does not depend on:
1) number of tons in both coils
2) cross sectional area of coils
3) magnetic permeability of medium between them
4) current flowing through the coils
Answer:
4) current flowing through the coils
Mutual inductance is a material property.
So, it does not depend on the current flowing through the coils.

Question 7.
In electromagnetic induction, the direction of induced current is such that it opposes the cause of its production. This is known as:
1) Kirchhoff’s law
2) Lenz’s law
3) Ampere’s law
4) Biot-Savart’s law
Answer:
2) Lenz’s law
Lenz’s law explains the negative sign in Faraday’s law equation, indicating that the system maintain the status of the magnetic flux.

Question 8.
A metal rod moves perpendicular to a uniform magnetic field. The emf induced is given by:
1) ε = B/v
2) ε = B l / v
3) ε = B l² v
4) ε = B v² l
Answer:
When a conductor of length l moves with a velocity v, perpendicular to magnetic field B then emf across the conductor is ε = BlN

Question 9.
A change in magnetic flux of 0.3 Wb in 0.1 s produces an average emf of:
1) 0.03 V
2) 3 V
3) 30 V
4) 300 V
Answer:
2) 3 V
Emf ε = \(\frac{d \phi}{d t}=\frac{0.3}{0.1}\) = 3V

Question 10.
If the flux linkage in a coil changes from 2 Wb to 8 Wb in 0.5 s, the average emf induced is:
1) 10V
2) 12 V
3) 6 V
4) 4 V
Answer:
2) 12 V
Emf e = \(\frac{d \phi}{d t}=\frac{6}{0.5}\) = 12V

Question 11.
The direction of induced current in a loop moving into a magnetic field is such that it:
1) enhances the magnetic field
2) opposes the relative motion
3) has no relation to the field
4) always flows clockwise
Answer:
2) opposes the relative motion
According to lenz’s law, the induced current will create a magnetic force that acts in the opposite direction of the loops motion to resist the change in flux.

II. Fill in the Blanks

Question 1.
According to Faraday’s law, the induced emf in a coil is directly proportional to the __________ of the magnetic flux.
Answer:
rate of change
From Faraday’s second law emf e = -N\(\frac{d \phi}{d t}\) ⇒ e ∝ \(\frac{d \phi}{d t}\) = Rate of change in flux.

Question 2.
Lenz’s law states that the direction of the induced current is such that it opposes the ___________ that causes it.
Answer:
change of magnetic flux
If the magnetic flux increases, the induced current creates a field to decrease it.
If it decreases, the current tries to increase it. So induced current always opposes the change of magnetic field

Question 3.
The SI unit of magnetic flux is ________
Answer:
Weber.
SI units of Magnetic Flux is Weber (Wb)

Question 4.
The back emf in an inductor is always in the direction opposite to the __________Answer:
original emf.
Back emf is opposite to the original emf.

Question 5.
The flux linkage through a coil of N-turns is proportional to the _________ passing through the coil.
Answer:
current
Magnetic flux is created by current. Flux linkage NΦ = LI

Question 6.
Self-inductance depends on the geometry of the coil, the number of turns, and the _________ of the core material.
Answer:
permeability
Self inductance of Solenoid L = \(\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)

Question 7.
The average induced emf is equal to the change in magnetic flux linkage divided by the ________ during which the change takes place.
Answer:
time
Emf E = \(\frac{d \phi}{d t}\). induced emf is the result of change in flux by the time duration of that charge.

Question 8.
The direction of induced current is determined by _______ law.
Answer:
Lenz
Faraday’s law gives us the magnitude and then Lenz’s law provides the direction.

III. One Word Answer Questions

Question 1.
Which law states that the induced emf is proportional to the rate of change of magnetic flux?
Answer:
Faraday’s law of induction.
It states that the induced emf in a circuit is directly proportional to the time rate of change of magnetic flux through that circuit.

Question 2.
Which law states that the induced current opposes the change in flux that produces it?
Answer:
Lenz’s law
It states that the direction of an induced current is always such as to oppose the change that causes it.

Question 3.
Give the expression for the emf induced between the ends of a metal conductor moving perpendicular to a uniform magnetic field.
Answer:
E = Blv. Here l is the length, v is the velocity, B is the uniform magnetic field

Question 4.
What will be the dimension of L/R, if L is inductance and R is resistance.
Answer:
The dimension of L/R is T
Inductance L = [ML²T^-2I^-2]; Resistance R = [ML²T^-3I^-2]
∴ \(\frac{\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{I}^{-2}}{\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-2}}\) = T

Question 5.
Why induced emf is also called back emf?
Answer:
According to Lenz’s law, the induced emf opposes the applied voltage.
Hence it is called back emf.

Question 6.
What is the working principle of a transformer?
Answer:
Transformer works on the principle of mutual inductance.

Question 7.
Why are resistance coils double wounded in opposite direction?
Answer:
To minimize/ avoid self inductance.

Question 8.
What is the change in self inductance of solenoid if iron core is inserted into solenoid?
Answer:
Inserting an iron core into a solenoid increases its self inductance L.

Question 9.
Write the expression for instantaneous value of emf in A.C generator.
Answer:
Instantaneous emf induced in an AC generator is E = NBAω sinωt
Here N is the number of turns; B is the magnetic induction.
A is the area of the coil; ω is the angular velocity

Question 10.
What is the unit of mutual inductance?
Answer:
SI unit of mutual inductance is Henry (H)

IV. Very Short Answer Questions

Question 1.
State Faraday’s law of electromagnetic induction.
Answer:
Faraday’s law: The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematical expression: Induced emf ε = –\(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
In the case of a coil of N turns, ε = – N \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)

Question 2.
State Lenz’s law.
Answer:
Lenz’s Law (Neumann’s law): The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux.
Faraday’s law of induction: ε = –\(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
The negative sign indicates Lenz’s law.

Question 3.
Define magnetic flux.
Answer:
Magnetic Flux is defined as the total number of magnetic field lines passing normally through j the given plane.
Formula: Φ_B = \(\vec{\mathrm{B}} . \vec{\mathrm{A}}\) = BA cos θ. Here θ is the angle between \(\vec{\mathrm{A}}\) and \(\vec{\mathrm{B}}\) .
Magnetic flux is a scalar. Its SI unit is Tm²

Question 4.
A magnet is moving towards a coil with a uniform speed ‘V’ as shown in figure (a) & (b). State the direction of induced current in the resistor ‘R’.

Answer:
According to Lenz’s law, the direction of induced current which opposes the cause of it.
a) If the magnet moves towards the coil, flux Increases.
Hence the induced current in the resistor R moves in anti-clockwise direction from Y to X.

b) If the magnet moves away from the coil, flux decreases.
Hence the induced current in the resistor R moves in clockwise direction from X to Y.

Question 5.
Write the expression for self-induced emf and explain the terms therein.
Answer:
The expression for self induced emf is ε = -L\(\frac{\mathrm{dI}}{\mathrm{dt}}\)
Here L is the self inductance, \(\frac{\mathrm{dI}}{\mathrm{dt}}\) is the rate of change of current.

Question 6.
What is self-induction?
Answer:
Self induction is the magnetic flux linked with the coil when unit current flows in it. (or)
When the current in a coil changes, an emf is induced in the coil opposing the change in the current through the coil. This phenomenon is called self induction.
∴ Φ ∝ I ⇒ Φ = LI (or) Emf = -L\(\frac{\mathrm{dI}}{\mathrm{dt}}\)

Question 7.
What is mutual induction?
Answer:
Mutual induction is defined as the flux linked with secondary coil when unit current flows through the primary coil.
(or)
When the current in one coil changes, an emf is induced in another coil kept near it. This phenomenon is called mutual inductance.
Φ ∝ I ⇒ Φ = MI (or) Emf = -M \(\frac{\mathrm{dI}}{\mathrm{dt}}\)

Question 8.
Why do birds fly off a high tension wire when the current is switched ‘ON’?
Answer:
When current is switched on, induced current flows through the body of the bird.
This induced current flows in the opposite direction through the wings of the bird.
Hence, the wings experience a force of mutual repulsion.
Thus, the wings spread and hence the birds fly due to electromagnetic induction.

Question 9.
What is the working principle of A.C. generator?
Answer:
AC generator works on the piinciple of electromagnetic induction.
The AC generator converts mechanical energy into electrical energy by rotating a coil with in a magnetic field, which changes the magnetic flux and induces an alternating emf.

Question 10.
Why is spark produced in the switch of a fan when it is switched oft?
Answer:
A fan motor contains coils of high inductance. When the fan is switched off, the current drops to zero rapidly. This sudden change induces a very high back emf (voltage) across the switch contacts.This high voltage is strong enough to ionize the air in the small gap of the switch, resulting in a visible spark.

Question 11.
A plot of magnetic flux (Φ) versus current (I) is shown in Figure for two inductors A and B. Which of the two has larger value of’seif induction. State the reason.

Answer:
In the given graph, slope of the line represents self inductance (L).
Slope = \(\frac{\phi}{I}\) = L
In the graph, slope of line OA is greater than the slope of the line OB.
Hence, the self inductance of A is greater than the self inductance of B. Thus L_A > L_B

V. Short Answer Questions

Question 1.
Explain the Faradays law of induction and Lenz’s law.
Answer:
Faraday’s law of Electromagnetic Induction : The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by ε = –\(\frac{d \phi_B}{d t}\)
In the case of a coil of N turns, ε = – N \(\frac{d \phi_B}{d t}\)
Lenz’s Law(Neumann’s law) : The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux.
Faraday’s law of induction is ε = –\(\frac{d \phi_B}{d t}\)
The negative sign in it indicates Lenz’s law.

Question 2.
Derive an expression for motional emf induced across the conductor moving perpendicular to uniform magnetic field.
Answer:
Induced emf due to motion of a conductor in a magnetic field:
Consider a rectangular conductor PQRS suspended in a uniform magnetic field B which is perpendicular to the plane of the system. The conductor PQ is free to move. Suppose, the conductor PQ is moved towards left with velocity v.

Let RQ = x, RS = l.
Then the magnetic flux enclosed by the loop PQRS is Φ_B = B l x
The induced emf is given by ε = – \(\frac{d \phi_B}{d t}\) = – \(\frac{d}{d t}\) (B l x) ⇒ ε = – B l \(\frac{d x}{d t}\)
But \(\frac{d x}{d t}\) = -v
Thus the induced emf in a conductor moving in a perpendicular and uniform magnetic field is given by ε = B l v

Question 3.
Obtain an expression for the mutual inductance of two long co-axial solenoids.
Answer:
Mutual Inductance is the phenomenon where a changing current in one coil induces an EMF in a nearby secondary coil.
Mutual induced emf ε = -M \(\frac{\mathrm{di}}{\mathrm{dt}}\), here M is mutual inductance of the coils.

Expression for mutual induction:
Consider two co-axial solenoids S₁, S₂ each of length l. Let A be the common cross¬section area of solenoids. Let N₁ and N₂ be the total number of turns in the solenoids S₁ and S₂ .
When current I₁ is passing through solenoid S₁, Magnetic field produced is B₁ = μ₀\(\frac{\mathrm{N}_1}{l}\)I₁ ………. (1)
Magnetic flux linked with each turn of solenoid S₂ is B₁ A = (μ₀\(\frac{\mathrm{N}_1}{l}\)I₁)A
∴ Total magnetic flux linked with N₂ turns of solenoid S₂ is, Φ₂ = (B₁ A)N₂
=* Φ₂ = (μ₀\(\frac{\mathrm{N}_1}{l}\)I₁) N₂ = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{I}_1 \mathrm{~A}}{l}\) …………. (2)
Now total magnetic flux linked with N₂ turns of solenoid S₂ is Φ₂ = M₁₂I₁ ………… (3)
where M₁₂ is mutual inductance of S₁ with respect to S₂
From eq (2) and eq (3) we get, M₁₂ I₁ = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{I}_1 \mathrm{~A}}{l}\) ⇒ M₁₂ = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~A}}{l}\) ………… (4)
Let n₁, n₂ be the number of turns per unit length of the solenoids S₁, S₂
∴ M₁₂ = μ₀n₁n₂lA ………….. (5) (∵ n₁ = \(\frac{\mathrm{N}_{1}}{l}\) , n₂ = \(\frac{\mathrm{N}_{2}}{l}\) )
Similarly, mutual inductance of solenoid S₂ with respect to S₁ is M₂₁ = μ₀n₁n₂lA ………. (6)
Comparing eq (5) and eq (6), we get M₂₁ = M₁₂
∴ Mutual inductance of two long coaxial solenoids, M = μ₀n₁n₂lA

Question 4.
Obtain an expression for the magnetic energy stored in a solenoid in terms of the magnetic field, area and length of the solenoid.
Answer:
Magnetic Energy stored in a Solenoid:
The energy required to build up the current in a solenoid is given by U = \(\frac{1}{2}\) LI² ……….. (1)
where L is the self inductance of the solenoid. 2
Self inductance of a solenoid is given by L = μ₀n² A l …………. (2)
here n is number of turns per unit length, A is area of cross section, l is length of the solenoid.
Substituting eq (2) in eq (1), we get U = \(\frac{1}{2}\) (μ₀ n² A l)I² ………….. (3)
Magnetic induction inside a solenoid is B = μ₀ n I ⇒ I = B/μ₀ n …………. (4)
Putting this in eq (3), we get U= \(\frac{1}{2}\) (μ₀ n² A l) \(\frac{B^2}{\mu_0{ }^2 n^2}\) = \(\frac{1}{2 \mu_0}\) B² Al
This is the expression for magnetic energy stored in a solenoid.

Question 5.
Explain briefly working of AC generator.
Answer:
AC Generator : AC generator converts mechanical energy into electrical energy.
Construction:
AC generator consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so that an emf is induced in the coil. The ends of the coil are connected to an external circuit by means of slip rings and brushes.

Working:
When the coil is rotated with a constant angular speed ω, θ = ωt.
where θ is the angle between vector area A and the magnetic field B at given time t.
The flux at time t is given by Φ_B = BA cos θ = BA cos ωt (∵ θ = ωt)
From Faraday’s law, the induced emf is ε = – N\(\frac{d \phi_B}{d t}\) = -N\(\frac{d}{d t}\)(BAcosωtt)
= -NBA \(\frac{d}{d t}\) (cos ωt)= -NBA (-sin ωt)(ω) = NBA ω sin ωt. dt
Thus the instantaneous emf is ε = NBA ω sin ωt.
Here, NBAω = ε_m = maximum emf.
∴ ε = ε_m sin ωt (∵ ε_m = NBAω)
Hence the voltage produced by the generator is Sinusoidal.

VI. Long Answer Questions

Question 1.
Outline the path-breaking experiments of Faraday and Henry and highlight the contributions of these experiments to our understanding of electromagnetism.
Answer:
Experiments of Faraday and Henry:
Experiment 1 (Current induced by magnet):
When a bar magnet is pushed towards a coil connected to a galvanometer, the galvanometer detected a current in the coil even though there is no battery in it.

Experiment 2(Current induced by current) :
Two coils are kept side by side. One of the coils is connected to a galvanometer without any battery.
The other coil is connected to a battery. When the current in the other coil is changed, a current is detected in the first coil.

Experiment 3 (Current induced by change in current) : Two coils are kept side by side. One of the coils is connected to a galvanometer without any battery. The other coil is connected to a battery with a key in series. When the key is closed or opened, a momentary current is detected in the first coil.

The common factor in all the above three experiments is that whenever the magnetic flux passing through the coil changes, a current is induced in it.
Basing on the results of these experiments, Faraday concluded that the induced emf in a coil is directly proportional to negative rate of change of magnetic flux passing through the coil.
∴ ε = –\(\frac{d \phi_B}{d t}\)

Question 2.
Describe the working of an AC generator with the aid of a simple diagram and necessary expressions.
Answer:
AC Generator : AC generator converts mechanical energy into electrical energy.
Construction:
AC generator consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so that an emf is induced in the coil. The ends of the coil are connected to an external circuit by means of slip rings and brushes.

Working:
When the coil is rotated with a constant angular speed ω, θ = ωt.
where θ is the angle between vector area A and the magnetic field B at given time t.
The flux at time t is given by Φ_B = BA cos θ = BA cos ωt (∵ θ = ωt)
From Faraday’s law, the induced emf is ε = – N\(\frac{d \phi_B}{d t}\) = -N\(\frac{d}{d t}\)(BAcosωtt)
= -NBA \(\frac{d}{d t}\) (cos ωt)= -NBA (-sin ωt)(ω) = NBA ω sin ωt. dt
Thus the instantaneous emf is ε = NBA ω sin ωt.
Here, NBAω = ε_m = maximum emf.
∴ ε = ε_m sin ωt (∵ ε_m = NBAω)
Hence the voltage produced by the generator is Sinusoidal.

Textual Solved Problems

Question 1.
A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.
Solution:
The angle θ made by the area vector of the coil with the magnetic field is 45°
The initial magnetic flux is Φ = BA cos θ = \(\frac{0.1 \times 10^{-2}}{\sqrt{2}}\) Wb
Final flux, θ_min = 0
The change in flux is brought about in 0.70s.
Induced emf ε = \(\frac{|\Delta \phi|}{\Delta \mathrm{t}}=\frac{|\phi-0|}{\Delta \mathrm{t}}=\frac{10^{-3}}{\sqrt{2} \times 0.7}\) = 1.0 mV
Magnitude of current I = \(\frac{\varepsilon}{R}=\frac{10^{-3}}{0.5}\) = 2 mA

Question 2.
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in he coil. Horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5 T.
Solution:
Initial flux through the coil, Φ_B(initial) = BA cos θ = 3.0 × 10^-5 × (π × 10^-2) × cos0° = 3π × 10^-7 Wb
Final flux after the rotation Φ_B(final) = 3.0 × 10^-5 × (π × 10^-2) × cos 180° = -3π × 10^-7 Wb
∴ Estimated value of the induced emf is,
ε = NA\(\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{500 \times\left(6 \pi \times 10^{-7}\right)}{0.25}\) = 3.8 × 10^-3 V = 3.8 mV
Magnitude of current I = \(\frac{\varepsilon}{R}=\frac{3.8 \times 10^{-3}}{2}\) = 1.9 mA

Question 3.
A wheel with 10 metallic spokes each 0.5 in long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_E
at a place. If H_E = 0.4 G at the placee what is the induced emf between the axle and the rim of the wheel? [Note that 1 G = 10^-4 T]
Solution:
Induced emf ε = \(\frac{1}{2}\)ωBR² = \(\frac{1}{2}\) × 4π × 0.4 × 10^-4 × (0.5)² = 6.28 × 10^-5 V
The number of spokes is immaterial because the emf is across the spokes are in parallel.

Exercise Problems

Question 1.
Find the magnetic flux linked with a rectangular coil of size 6 cm × 8 cm placed at right j angles to a magnetic field of 0.5 wb/ m²
Solution:
Given Length l = 6cm = 0.06m, width b = 8cm = 0.08m, Magnetic induction B = 0.5 Wb/m².
Area A = l x b = 0.06 × 0.08 = 0.0048m²
If the coil is perpendicular to the field, the normal is parallel to the field. So, 0 = 0°.
Magnetic flux Φ = BA cos θ = (0.5) (0.0048) cos0° = 0.0024 = 2.4 × 10^-3 Wb

Question 2.
A small metal piece of metal wire is dragged across the gap between pole pieces of a magnet in 0.5 s. The magnetic flux between the pole pieces is known to be 8 × 10^-4 wb. Estimate the emf induced in the wire.
Solution:
Change in magnetic flux ∆Φ = 8 × 10^-4 wb, Time interval ∆t = 0.5 s
∴ Induced emf ε = \(\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{8 \times 10^{-4}}{0.5}\) = 16 × 10^-4 = 1.6 × 10^-3V

Question 3.
A circular coil of area 300 cm2 and 25 turns rotates about its vertical diameter with an angular speed of 40 rad/s in a uniform horizontal magnetic field of magnitude 0.05T. Obtain the maximum voltage induced in the coil.
Solution:
Given number of turns N = 25, Area A = 300cm2 = 0.03 m², Angular speed ω = 40 rad/s
Induction B = 0.05T
The maximum induced voltage E_max = NBAω = 25 × 0.05 × 0.03 × 40 = 1.5V

Question 4.
An air craft with a wing span of 40m flies with a speed of 1080 Km/h in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth’s magnetic field is 1.75 × 10^-5 T . Find the emf that develops between the tips of the wings.
Solution:
The speed of air craft v= 1080 km/h = \(\frac{1080 \times 5}{18}\) = 300m / s; wing span L = 40m
Vertical component of magnetic field B_v = 1.75 × 10^-5 T
∴ Induced emf ε = B_vVL = 1.75 × 10^-5 × 300 × 40= 0.21V

Question 5.
If a rate of change of current of 4 A/s induces an emf of 20 mV in a solenoid. What is the self-inductance of the solenoid.
Solution:
Rate of change of current \(\frac{\mathrm{dI}}{\mathrm{dt}}\) = 4A/s; Induced emf ε = 20mV = 20 × 10^-3 V
Formula: ε = L\(\frac{\mathrm{dI}}{\mathrm{dt}}\) ⇒ L = \(\frac{\varepsilon}{(\mathrm{di} / \mathrm{dt})}=\frac{20 \times 10^{-3}}{4}\) = 5 × 10^-3V = 5 mH

Question 6.
The mutual inductance of two coaxial coils is 2H. The current in one coil is changed uniformly from zero to 0.5A in 100 ms. Find the (a) change in magnetic flux through the other coil (b) emf induced in the other coil during the change.
Solution:
Given Mutual inductance M = 2H, Change in current ∆I = 0.5 – 0 = 0.5A,
Time interval ∆t = 100ms = 100 × 10^-3 s = 0.1 s
(a) Change in magnetic flux ∆Φ = M ∆I = 2 x 0.5 = 1 Wb.
(b) Emf induced in the other coil ε = M\(\frac{\mathrm{M}}{\mathrm{At}}\) = 2 × \(\frac{0.5}{0.1}=\) = 10V

Question 7.
If the self-inductance of an air core inductor increases from 0.01 m H to 10 m H on introducing an iron core into it. What is the relative permeability of the core used.
Solution:
L_air = 0.01 mH; L_core = 10mH
Relative permeability μ_r = \(\frac{\mathrm{L}_{\text {core }}}{\mathrm{L}_{\text {air }}}=\frac{10}{0.01}\) = 1000.

Question 8.
A metallic wire 1m in length is moving normally across a magnetic field of 0.1 T with a speed of 5 m/s. Find the emf induced between the ends of the wire.
Solution:
Length of the wire L = 1 m, Magnetic induction B = 0.1 T
Speed v = 5m/s, Angle θ = 90° (the wire moves normal to the field)
The induced emf E = BLV sinθ = 0.1 × 1 × 5 × sin90° = 0.5 V

Objective Questions

Question 1.
A circular disc of radius 0.2 meter is placed in a uniform magnetic Odd of induction \(\frac{1}{\pi}\left(\frac{W b}{m^2}\right)\) in such a way that its axis makes an angle of 60° with \(\vec { B }\). The magnetic flux linked with the disc is
1) 0.08 Wb
2) 0.01 Wb
3) 0.02 Wb
4) 0.06 Wb
Answer:
3) 0.02 Wb

Question 2.
A 800 turn coil of effective area 0.05 m2 is kept perpendicular to a magnetic field 5× 10^-5, T. When the plane of the coil is rotated by 90° around any of its coplanar axis in 0.1 s, the emf induced in the coil will be
1) 0.02 V
2) 2 V
3) 0.2 V
4) 2 × 10^-3 V
Answer:
1) 0.02 V

Question 3.
A coil of resistance 400 Ω is placed in a magnetic field. If the magnetic flux Φ (Wb) linked with the coil varies with time t (sec) as Φ = 50t² + 4. The current in the coil at t = 2 sec is
1) 0.5 A
2) 0.1 A
3) 2 A
4) 1 A
Answer:
1) 0.5 A

Question 4.
A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is
1) 4.8πµ V
2) 0.8πµV
3) 1.67πµ V
4) 3.47πµ V
Answer:
4) 3.47πµ V

Question 5.
A rectangular coil of 20 turns and area of cross-section 25 sq. cm has a resistance of 100 Ω. If a magnetic field which is perpendicular to the plane of coil changes at a rate of 1000 tesla per second, the current in the coil is
1) 1 A
2) 50 A
3) 0.5 A
4) 5 A
Answer:
3) 0.5 A

Question 6.
A magnetic field of 2 × 10^-2 T acts at right angles to a coil of area 100 cm2, with 50 turns. The average e.m.f. induced in the coil is 0.1 V, when it is removed from the field in t sec. The value of t is
1) 10 s
2) 0.1s
3) 0.01s
4) 1 s
Answer:
2) 0.1s

Question 7.
A metal ring is held horizontally and bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet is
1) more than g
2) equal to g
3) less than g
4) either (1) or (3)
Answer:
3) less than g

Question 8.
Faraday’s laws are consequence of conservation of
1) energy
2) energy and magnetic field
3) charge
4) magnetic field
Answer:
1) energy

Question 9.
A straight line conductor of length 0.4 m is moved with a speed of 7 m/s perpendicular to a magnetic field of intensity 0.9 Wb/’m². The induced e.m.f. across the conductor is
1) 5.04 V
2) 25.2 V
3) 1.26 V
4) 2.52 V
Answer:
4) 2.52 V

Question 10.
The total charge, induced in a conducting loop when it is moved in magnetic field depends on
1) the rate of change of magnetic flux
2) initial magnetic flux only
3) the total change in magnetic flux
4) final magnetic flux only.
Answer:
3) the total change in magnetic flux

Question 11.
In which of the following devices, the eddy current effect is not used?
1) electric heater
2) induction furnace
3) magnetic braking in train
4) electromagnet
Answer:
1) electric heater

Question 12.
Eddy currents are produced when
1) a metal is kept in varying magnetic field
2) a metal is kept in steady magnetic field
3) a circular coil is placed in a magnetic field
4) current is passed through a circular coil
Answer:
1) a metal is kept in varying magnetic field

Question 13.
Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ >> R₂. the mutual inductance M between them will be directly proportional to
1) \(\frac{\mathrm{R}_2^2}{\mathrm{R}_1}\)
2) \(\frac{R_1}{R_2}\)
3) \(\frac{R_2}{R_1}\)
4) \(\frac{R_1^2}{R_2}\)
Answer:
1) \(\frac{\mathrm{R}_2^2}{\mathrm{R}_1}\)

Question 14.
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
1) 0.138 H
2) 138.88 H
3) 1.389 H
4) 13.89 H
Answer:
4) 13.89 H

Question 15.
Two coils of self inductance 2 mH and 8 mil are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is
1) 16 mH
2) 10 mH
3) 6 mH
4) 4 mH
Answer:
4) 4 mH

Question 16.
For a coil having L = 2 mH, current flow through it is I = t²e^-1 then, the time at which emf becomes zero
1) 2 sec
2) 1 sec
3) 4 sec
4) 3 sec
Answer:
2) 1 sec

Question 17.
Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I = I_o sinωt, where I_o = 10 A and ω = 100π rad/sec. The maximum value of e.m.f. in the second coil is
1) π
2) 5 π
3) 2π
4) 4π
Answer:
2) 5 π

Question 18.
What is the self-inductance of a coil which produces 5 V when the current changes from 3 ampere to 2 ampere in one millisecond?
1) 5000 henry
2) 5 milli-henry
3) 50 henry
4) 5 henry
Answer:
2) 5 milli-henry

Question 19.
If the number of turns per unit length of a coil of solenoid is doubled, the self¬inductance of the solenoid will
1) remain unchanged
2) be halved
3) be doubled
4) become four times
Answer:
4) become four times

Question 20.
An inductor may store energy in
1) its electric field
2) its coils
3) its magnetic field
4) both in electric and magnetic fields
Answer:
3) its magnetic field

Question 21.
A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is
1) four times per revolution
2) six times per revolution
3) once per revolution
4) twice per revolution
Answer:
4) twice per revolution

Regular practice with AP Inter 2nd Year Physics Study Material Chapter 5 Magnetism and Matter Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Physics 5th Lesson Magnetism and Matter Questions and Answers

I. Multiple Choice Questions

Question 1.
The magnetic field lines of the bar magnet looks similar to the magnetic field lines of
1) a current carrying circular coil
2) a straight long current carrying wire
3) a long solenoid
4) a plane current sheet
Answer:
3) a long solenoid
A long solenoid produces a magnetic field that is similar to that of a bar magnet, with field lines forming continuous loops through the centre and spreading out of the poles.

Question 2.
A bar magnet of magnetic moment m is placed in a uniform magnetic field B. Its potential energy is zero when the angle between magnetic moment and magnetic field is ….
1) 0°
2) 90°
3) 180°
4) 45°
Answer:
2) 90°
Potential energy U = -MB cosθ = 0 ⇒ cosθ = 0 ⇒ θ = 90°

Question 3.
Which of the following is not true for the magnetic field lines?
1) The magnetic lines are open curves
2) The magnetic lines never intersect
3) The direction of magnetic induction is tangential to the field line at a point
4) The magnetic field is strong where the field lines are denser
Answer:
1) The magnetic lines are open curves
Magnetic lines always form closed loops. Hence open curves are not possible.

Question 4.
The magnetic field induction and magnetisation in a paramagnetic material are \(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{M}}\). Then the magnetic field intensity is ________
1) \(\overrightarrow{\mathrm{H}}=\mu_0(\overrightarrow{\mathrm{~B}}+\overrightarrow{\mathrm{M}})\)
2) \(\overrightarrow{\mathrm{H}}=\frac{1}{\mu_0}(\overrightarrow{\mathrm{~B}}+\overrightarrow{\mathrm{M}})\)
3) \(\overrightarrow{\mathrm{H}}=\frac{\overrightarrow{\mathrm{M}}}{\mu_0}-\overrightarrow{\mathrm{B}}\)
4) \(\overrightarrow{\mathrm{H}}=\frac{\overrightarrow{\mathrm{B}}}{\mu_0}-\overrightarrow{\mathrm{M}}\)
Answer:
4) \(\overrightarrow{\mathrm{H}}=\frac{\overrightarrow{\mathrm{B}}}{\mu_0}-\overrightarrow{\mathrm{M}}\)
The relation between B, H & M is B = μ₀(H + M)
⇒ H + M = \(\frac{\mathrm{B}}{\mu_0}\)
∴ H = \(\frac{\mathrm{B}}{\mu_0}\) – M

Question 5.
The universal character of every material is
1) diamagnetic
2) paramagnetic
3) ferromagnetic
4) ferrimagnetic
Answer:
1) diamagnetic
Diamagnetism is a fundamental property present in all atoms and molecules.
It arises Iromthe orbital motion of atoms.

Question 6.
Which of the following materials exhibits paramagnetic property?
1) Iron
2) Cobalt
3) Aluminium
4) Nickel
Answer:
3) Aluminium
Aluminium is paramagnetic. Iron, Nickel and cobalt are ferromagnetic.

Question 7.
A freely suspended magnet align in direction.
1) South-West
2) East-West
3) North-South
4) North-East
Answer:
3) North-South
Due to earth’s magnetic field, a suspended magnet always aligns, it self along geographic north and south poles.

Question 8.
Which one among the following materials exhibits higher magnetic susceptibility?
1) Ferromagnetic
2) Paramagnetic
3) Diamagnetic
4) Both Paramagnetic and Diamagnetic
Answer:
1) Ferromagnetic
Ferromagnetic materials have very high and positive susceptibility (𝜒).

II. Fill in the Blanks

Question 1.
The minimum potential energy of the bar magnet of magnetic moment m placed in a uniform magnetic field B is ________.
Answer:
-mB.
When the magnet is aligned parallel to the field, θ = 0° or 180°
When θ = 180° , Potential energy U= MB cos θ = MB cos 180° = MB(-1 )= -MB

Question 2.
The SI unit of magnetisation (M) is _________
Answer:
ampere/metre.
SI Units of M=Ampere/metre (or) A/m

Question 3.
The SI unit of magnetic field intensity (H) is _________
Answer:
ampere/metre.
SI Units of H = Ampere/ metre (or) A/m

Question 4.
The magnetic material which tries move from strong field region to weak field region is known as _________
Answer:
diamagnetic.
Diamagnetic materials are repelled from strong field to weak field.

Question 5.
The magnetic material whose susceptibility is positive but small is ________
Answer:
paramagnetic.
Paramagnetic materials have a small positive susceptibility.

Question 6.
There is __________ kind of force when the north poles of two magnets are brought close together.
Answer:
repulsive
Repulsive force exists when the north poles (like poles) are brought close together.

Question 7.
A long magnet of length 2L and magnetic moment M and pole strength m units is broken into two pieces at the middle then magnetic moment of the piece is _______ than the original magnet.
Answer:
less
Magnetic moment M = pole strength × length ⇒ M = m2l. When the magnet is broken into two pieces at the middle, length becomes half.
So, M’ = m\(\left[\frac{2 l}{2}\right]=\frac{2 l \mathrm{~m}}{2}=\frac{\mathrm{M}}{2}\)

Question 8.
The sign of susceptibility of the diamagnetic material is _________.
Answer:
negative
For diamagnetic material, Susceptibility is negative. Induced magnetic moment of these materials is opposite direction of applied field.

Question 9.
Domains are the characteristic feature of _________ magnetic materials
Answer:
ferro
Domains are the characteristic feature of ferromagnetic materials.

Question 10.
The net magnetic flux passing through any closed surface around bar magnet is _______
Answer:
zero.
The net magnet flux passing through any closed surface around bar magnet is zero.
Based as Gauss’s law for magnetism, the magnetic monopoles do not exists.

III. One Word Answer Questions

Question 1.
What happens to magnetic moment of a magnet, if it is cut into two pieces along its length?
Answer:
Magnetic moment becomes half of the original (M’=M/2) ‘
Magnetic moment M = pole strength × length ⇒ M = m2l.
When the magnet is broken into two pieces, length becomes half.
So, M’ = m\(\left[\frac{2 l}{2}\right]=\frac{2 l \mathrm{~m}}{2}=\frac{\mathrm{M}}{2}\)

Question 2.
What is the magnetic field induction at a distance of r from the center of the bar magnet of magnetic moment m on its axis?
Answer:
Along the axial line, for short magnet B_axial = \(\frac{\mu_0}{4} \frac{2 M}{3}\). M = magnetic moment, r = distance

Question 3.
What is the magnetic field induction at a distance of r from the center of the bar magnet of magnetic moment rn on its bisector line (equatorial line)?
Answer:
Along the equatorial line, for short magnet B_eq = \(\frac{\mu_0}{4} \frac{M}{3}\). M = magnetic moment, r = distance

Question 4.
What is the analogy to permittivity ε₀ free space in magnetic field?
Answer:
Permeability of free space (μ₀) [μ₀ = 4π × 10^-7 H/m.]

Question 5.
Write a relation between susceptibility and relative permeability.
Answer:
μ_r = 1 + 𝜒 is a relation between relative permeability (μ₀) and magnetic susceptibility (𝜒)

Question 6.
Which law confirms that no mono magnetic poles exist?
Answer:
Gauss’s law for magnetism is the law that confirms that no mono magnetic poles exist.

Question 7.
What is magnetization of a sample?
Answer:
Magnetisation M of a sample is the net magnetic moment per unit volume.

Question 8.
How many atoms does a magnetic domain roughly contain?
Answer:
10¹¹ to 10¹⁸ atoms will be there in the magnetic domain of ferromagnetic material.

Question 9.
Write down (he relation between magnetic induction (II) and magnetic Intensity (II).
Answer:
B = μH(or)B = μ₀μ_rH
This is the relation between magnetic induction (B) and magnetic field intensity (H).

IV. Very Short Answer Questions

Question 1.
State Gauss law in magnetism and write it its mathematical expression.
Answer:
Gauss law: The net magnetic flux (Φ) through any closed surface is equal to zero.
Mathematical expression: \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{\mathrm{~S}}\) = 0
Here \(\vec{\mathrm{B}}\) is the magnetic vector, and \(\mathrm{~d} \vec{\mathrm{~S}}\) is surface vector area.

Question 2.
Define magnetisation of a material and write its formula.
Answer:
Magnetisation (M) is defined as the net magnetic dipole moment per unit volume of the material.
Formula: M = \(\frac{m_{\text {net }}}{V}\)

Question 3.
Define magnetic susceptibility and relative permeability.
Answer:
a) Magnetic Susceptibility (𝜒) is the ratio between magnetisation (M) and magnetic intensity (H) Formula: 𝜒 = \(\frac{\mathrm{M}}{\mathrm{H}}\). It has no units. It determines how easily a material can be magnetized. H.
b) Relative permeability (μ_r) is the ratio of the magnetic permeability of a specific medium (μ) to the magnetic permeability of free space (μ₀) μ_r = μ/μ₀

Question 4.
What is the magnetic moment associated with a solenoid?
Answer:
Magnetic moment of a solenoid is \(\overline{\mathrm{M}}=\mathrm{NI} \overline{\mathrm{~A}}\). Here N is the number of turns, I is the current passing through it and \(\overline{\mathrm{~A}}\) is the area of cross section.

Question 5.
Magnetic lines form continuous closed loops. Why?
Answer:
The magnetic field lines of a magnet or a solenoid form continuous closed loops because the magnetic poles N and S always exist together in pairs.

Question 6.
Classify the following materials with regard to magnetism.
Iron, Platinum, Copper, Magnesium.
Answer:
Ferromagnetic material: Iron
Paramagnetic material: Platinum, Magnesium
Diamagnetic material; Copper

Question 7.
Magnetic lines of force don’t intersect. Why?
Answer:
Any two magnetic field Lines do not intersect each other because if they do, it means that at the point of intersection it shows two different directions which is impossible.

V. Short Answer Questions

Question 1.
Mention the properties of magnetic field lines.
Answer:
Properties of magnetic field lines :

1. The magnetic field lines of a magnet form continuous closed loops.
2. The magnetic line of force starts from a north pole and ends at south pole outside the magnet and will be from south pole to north pole inside the magnet.
3. At a given point of the field line, the tangent represents the direction of the field at that point.
4. The magnetic field lines don’t intersect. If they intersect, magnetic field at that point will have ‘ two directions which is not possible.
5. The magnetic lines of force experience longitudinal strain. Due to this property unlike poles attract each other.
6. The magnetic lines of force exert lateral pressure. Due to this property like poles repel each other.

Question 2.
State Gauss’ law-in magnetism. Show that how this law proved the non-existence of magnetic monopoles.
Answer:
Gauss’ Law for Magnetism: Gauss’ Law for magnetism states that the net magnetic flux through any closed surface is always zero.
Gauss law for magnetism is \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{\mathrm{~S}}\) = 0
Explanation: The above equation states that the net magnetic flux through any closed surface is zero. Magnetic filed lines always form closed continous loops. They do not begin or end at any point, unlike electric field lines.
Non-existence of Magnetic Monopoles: Since the net flux through a closed surface is zero there is no isolated magnetic poles (monOpoles) in nature. Magnetic poles always exist in pairs (North & south)
Conclusion: Gauss law in Magnetism proves that mangetic monopoles do not exist.

Question 3.
Explain the properties of diamagnetic materials.
Answer:
Properties of diamagnetic substances: .

1. Diamagnetic substances are weakly magnetised in opposite direction to that of magnetising field
   Ex: Cu, Ag; Au, H₂O
2. They are feebly repelled by a magnet.
3. They move from higher intensity to lower intensity in a non-uniform magnetic field.
4. They are independent of temperature.
5. Their relative permeability μ_r < 1, (𝜒 < 0, 𝜒 is small and negative) .
6. Resultant magnetic moment of its atom is zero, in the absence of external field.

Question 4.
Explain the properties of paramagnetic materials.
Answer:
Properties of Paramagnetic substances:

1. Paramagnetic substances are weakly magnetised in the direction of magnetising field.
   Ex: Al, Na, Pt.
2. They are feebly attracted by a magnet.
3. They move from lower intensity to higher intensity in a non-uniform magnetic field.
4. They are dependent on temperature.
5. Their relative permeability μ_r > 1., (𝜒 > 0, 𝜒 is small and positive)
6. Resultant magnetic moment of its atom is non-zero, due to unpaired electrons.

Question 5.
How the existence of domains, explain the behavior of ferromagnetic material?
Answer:
Ferromagnetic materials are the substances which are strongly attracted by magnets.
Their atomic dipoles align spontaneously to form strong clusters called domains.

1. In ferromagnetic materials the atoms form innumerable small effective regions called “domains”.
2. The size of domain varies from about 10^-6 m to 10^-3m.
3. Each domain contains 10¹⁷ to 10²¹ atomic magnets aligned in the same direction.
4. In an external magnetic field, the resultant magnetism increases due to reorientation of all – domains into the direction of field.
5. In the absence of external field the domains are oriented randomly, so that net magnetic moment is zero.

Question 6.
Define magnetisation of a sample, magnetic induction and susceptibility. How magnetic induction is related to magnetic intensity?
Answer:
a) Magnetisation (M) of a sample is defined as the net magnetic moment per unit volume.
Formula: M = \(\frac{\mathrm{m}_{\text {net }}}{\mathrm{V}}\) unit: A m^-1.
b) Magnetic induction(B) is defined as the total number of magnetic lines of force passing normally through a unit area of a substance. SI unit: Tesla (T)
c) Magnetic Susceptibility (x) is defined as the ratio between magnetisation (M) and magnetic intensity (H). Formula: 𝜒 =\(\frac{\mathrm{M}}{\mathrm{H}}\). It has no units

Relation between (B) and (H):
In a material of a susceptibility 𝜒_m, magnetic induction B is proportional to the magnetic intensity H Thus B ∝ H ⇒ B = μH; Here μ = permeability of the substance.
In free space, B = μ₀H
The total magnetic induction B = B (applied field) + B (Magnetization)
∴ B = μ₀H + μ₀H 𝜒_m = μ₀H( 1 + 𝜒_m)
= μ₀(1 + 𝜒_m)H = μ₀(μ_r)H [∵ 1 + 𝜒_m = μ_r]
∴ B = μH [∵ μ₀(μ_r) = μ]

VI. Long Answer Questions

Question 1.
Prove that a bar magnet and solenoid produce similar fields.
Answer:
Consider a solenoid of length 21 and radius R, containing n turns per unit length carrying current I.
Let P be a point on its axial line at a distance r from its centre O. Let us consider an element of thickness dx at a distance x from its centre.

The number of turns on the element will be n dx.
We know the magnetic field on the axis of the coil B = \(\frac{\mu_0 \mathrm{NIR}^2}{2\left(x^2+\mathrm{R}^2\right)^{3 / 2}}\)
Now by taking N = ndx, replacing x by (r – x) and taking differential, we get
dB = \(\frac{\mu_0 n d x \mathrm{IR}^2}{2\left[(r-x)^2+R^2\right]^{3 / 2}}\)
To get total magnetic field, we integrate the above equation between the limits -l to +l
B = \(\frac{\mu_0 \mathrm{nIR}^2}{2} \int_{-l}^l \cdot \frac{\mathrm{dx}}{\left[(\mathrm{r}-\mathrm{x})^2+\mathrm{R}^2\right]^{3 / 2}}\)
When x<2 +R²]^3/2 = r³
∴ B = \(\frac{\mu_0 \mathrm{nIR}^2}{2} \int_{-l}^l \frac{\mathrm{dx}}{\mathrm{r}^3}=\frac{\mu_0 \mathrm{nIR}^2}{2 \mathrm{r}^3} \int_{-l}^l \mathrm{dx}=\frac{\mu_0 \mathrm{nIR}^2}{2 \mathrm{r}^3}(2 l)=\frac{\mu_0 \mathrm{n}(2 l) \mathrm{I}\left(\pi \mathrm{R}^2\right)}{2 \pi \mathrm{r}^3}\)
But n (2l)I(πR²) = m = magnetic moment of the solenoid.
∴ B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}\)
This is the expression for the magnetic field on the axial field of a solenoid.
The axial field of a bar magnet is also given by same equation B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}\)
Thus, a bar magnet and a solenoid produce similar magnetic fields.
Hence, a solenoid behaves like a magnetic dipole similar to a bar magnet.

Question 2.
Explain the properties of dia, para and ferro magnetic materials.
Answer:
Properties of dia Magnetic substances:

1. Diamagnetic substances are weakly magnetised in opposite direction to that of magnetising field Ex: Cu, Ag, Au, H₂O
2. They are feebly repelled by a magnet.
3. They move from higher intensity to lower intensity in a non-uniform magnetic field.
4. They are independent of temperature.
5. Their relative permeability μ_r < 1, (𝜒 < 0, 𝜒 is small and negative)
6. Resultant magnetic moment of its atom is zero, in the absence of external field.

Properties of Paramagnetic substances:

1. Paramagnetic substances are feebly magnetised in the direction of magnetising field.
   Ex: Ai, Na, Pt, etc
2. They are feebly attracted by a magnet.
3. They move from lower intensity to higher intensity in a non-uniform magnetic field.
4. They are dependent on temperature.
5. Their relative permeability (μ_r > 1., (𝜒 > 0, 𝜒 is small and positive)
6. Resultant magnetic moment of its atom is non-zero, due to unpaired electrons.

Properties of Ferromagnetic substances:

1. Ferromagnetic substances are strongly magnetised in the direction of magnetising field.
   Ex:Iron, cobalt, nickel etc.
2. They are strongly attracted by a magnet.
3. They move from lower intensity to higher intensity in a non-uniform magnetic field.
4. A ferromagnetic substance turns into a paramagnetic at a temperature called Curie temperature.
5. Their relative permeability μ_r >> 1 (10 to 10⁴)
6. In ferromagnetic substances, domains having high, non-zero magnetic moments are formed.

Textual Solved Problems

Question 1.
(a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point?
(b) If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified?
(c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another element of the same wire?
(d) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero?
Solution:
(a)No. The magnetic force is always normal to B (remember magnetic force = qv B). It is misleading to call magnetic field lines as lines of force.

(b) Gauss’s law of magnetism states that die flux of B through any closed surface is always zero \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{\mathrm{~S}}\) = 0. If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to Gauss’s law of electrostatics, \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{\mathrm{~S}}\) = μ₀q_m,
here qm is the (monopole) magnetic charge enclosed by S .]

(c) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. .
(For the special case of a straight wire, this force is zero.)

(d) Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.

Question 2.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current IM.
Solution:
(a) The field His dependent of the material of the core, and is
H = nI = 1000 × 2.0 = 2 × 10³ A/m.

(b) The magnetic field B is given by B = μ_rμ₀H = 400 × 4π × 10^-7 × 2 × 10³ = 1.0T

(c) Magnetisation is given by M = (B – μ₀H)/μ₀
= (μ_rμ₀H – μ₀H) / μ₀ = (μ_r – 1)H – 399 × H ≅ 8 × 10⁵ A / m

(d) The magnetising current IM is die additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μ_rn(I + I_M)
Using I = 2A, B = 1 T, we get I_M = 794 A.

Exercise Problems

Question 1.
The magnetic moment of a bar magnet is 8 × 10³ Am². What is the magnetic field induction at a distance 4 cm oil its equatorial line?
Solution:
Given Magnetic moment M = 8 × 10³ Am², distance r = 4cm = 0.04m.
Magnetic induction at a point in equatorial line B = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^3}\)
= \(\frac{4 \pi \times 10^{-7}}{4 \pi} \frac{8 \times 10^3}{(0.04)^3}\) = 0.125 × 10² = 12.5
∴ Magnetic induction B = 12.5 T

Question 2.
What is the potential energy of the magnetic moment if it is placed in a uniform magnetic field such that its magnetic moment points opposite to the magnetic field?
Solution:
Potential Energy (U) = –\(\vec{\mathrm{M}} \cdot \vec{\mathrm{~B}}\) = -MBcosθ, θ is the angle between \(\vec{\mathrm{M}}\) and \(\vec{\mathrm{B}}\)
When magnetic moment points opposite to the field, θ = 180°
∴ U = -MB cos 180° = – MB (- 1) = MB
∴ The potential energy of a magnetic moment M, placed in a field B is U = MB

Question 3.
The magnetic susceptibility of a material is 0.1. Then what is its relative permeability?
Solution:
Magnetic susceptibility 𝜒 = 0.1
∴ Relative permeability μ_r = 1 + 𝜒 = 1 + 0.1 = 1.1

Question 4.
A short bar magnet placed with its axis at 30″ with a uniform external magnetic field of 0.45T experiences a torque of magnitude equal to 4.5 xlO-2 J . What is the magnitude of magnetic moment of the magnet?
Solution:
Magnetic induction B = 0.45 T, Angle θ = 30°, Torque 𝜏 = 4.5 × 10²J
Formula: 𝜏 = MB sinθ
Magnetic moment M = \(\frac{\tau}{B \sin \theta}=\frac{4.5 \times 10^{-2}}{0.45 \times \sin 30^{\circ}}=\frac{0.0045}{0.45 \times 0.5}\) = 0.2 JT^-1

Objective Questions

Question 1.
A bar magnet of magnetic moment M is placed at right angles to a magnetic induction B. If a force F is experienced by each pole of the magnet, the length of the magnet will be Q]
1) MB/F
2) BF/M
3) MF/B
4) F/MB
Answer:
1) MB/F

Question 2.
A magnetic needle suspended parallel to a magnetic field requires \(\sqrt{3}\)J of work to turn it through 60°. The torque needed to maintain the needle in this position will be
1) 2\(\sqrt{3}\)J
2) 3J
3) \(\sqrt{3}\)J
4) 3/2 J
Answer:
2) 3J

Question 3.
A bar magnet having a magnetic moment of 2 × 10^-4JT^-1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 10^-4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is
1) 12
2) 6J
3) 2J
4) 0.6J
Answer:
2) 6J

Question 4.
A bar magnet of magnetic moment \(\vec{\mathrm{M}}\), is placed in a magnetic field of induction \(\vec{\mathrm{B}}\). The torque exerted on it is
1) \(\vec{\mathrm{M}} \times \vec{\mathrm{B}}\)
2) \(-\vec{\mathrm{M}} \cdot \vec{\mathrm{~B}}\)
3) \(\vec{\mathrm{M}} \cdot \vec{\mathrm{~B}}\)
4) \(-\vec{\mathrm{B}} \cdot \vec{\mathrm{~M}}\)
Answer:
1) \(\vec{\mathrm{M}} \times \vec{\mathrm{B}}\), 4) \(-\vec{\mathrm{B}} \cdot \vec{\mathrm{~M}}\)

Question 5.
A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of each part will be
1) M
2) 2M
3) zero
4) 0.5M
Answer:
4) 0.5M

Question 6.
Tangent galvanometer is used to measure
1) potential difference
2) current
3) resistance
4) charge.
Answer:
2) current

Question 7.
An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m’. The permeability of the material of the rod is μ₀ = 4π × 10^-7 T m A^-1)
1) 2.4π × 10^-4 T m A^-1
2) 8.0 × 10^-5 T m A^-1
3) 2.4π × 10^-5 T m A^-1
4) 2.4π × 10^-7 T m A^-1
Answer:
1) 2.4π × 10^-4 T m A^-1

Question 8.
The magnetic moment of a diamagnetic atom is
1) much greater than one
2) 1
3) between zero and one
4) equal to zero
Answer:
4) equal to zero

Question 9.
Curie temperature above which
1) paramagnetic material becomes ferromagnetic material
2) ferromagnetic material becomes diamagnetic material
3) ferromagnetic material becomes paramagnetic material
4) paramagnetic material becomes diamagnetic material
Answer:
3) ferromagnetic material becomes paramagnetic material

Question 10.
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show
1) anti ferromagnetism
2) no magnetic property
3) diamagnetism
4) paramagnetism
Answer:
4) paramagnetism

Question 11.
According to Curie’s law, the magnetic susceptibility of a substance at an absolute temperature T is proportional to
1) 1/T
2) T
3) 1/T²
4) T²
Answer:
1) 1/T

Question 12.
Among which the magnetic susceptibility does not depend on the temperature?
1) Diamagnetism
2) Paramagnetism
3) Ferromagnetism
4) Ferrite
Answer:
1) Diamagnetism

Regular practice with AP Inter 2nd Year Physics Study Material Chapter 4 Moving Charges and Magnetism Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Physics 4th Lesson Moving Charges and Magnetism Questions and Answers

I. Multiple Choice Questions

Question 1.
A positron, moving vertically up from the ground, experiences a magnetic force by a magnetic field, directed due east. The magnetic force on it points due
1) South
2) North
3) West
4) South-east
Answer:
2) North
For a charged particle,, \(\vec{F}=q(\vec{v} \times \vec{B})\). Since a positron is a positive charge, the force
direction is given directly by the right-hand rule.
Velocity \(\vec{v}\) → vertically upward; Magnetic field \(\vec{B}\) → due east
Using right-hand rule: Upward × East = North

Question 2.
An electron moves through a uniform magnetic field B at a velocity v at time t. Then the magnetic force experienced by it is
1) parallel to \(\vec{v}\) only
2) perpendicular to \(\vec{v}\) × \(\vec{B}\)
3) opposite to \(\vec{B}\) only
4) opposite to \(\vec{v}\) × \(\vec{B}\)
Answer:
4) opposite to \(\vec{v}\) × \(\vec{B}\)
From the equation, \(\vec{F}=q(\vec{v} \times \vec{B})\), due to the negative charge of electron, the direction of force is exactly opposite to the direction of the vector cross product \(\vec{v} \times \vec{B}\)

Question 3.
A straight conducting wire, carrying an electric current of 0.1 A, is kept perpendicular to the uniform magnetic field of 1 T. The magnetic force per meter experienced by the wire is ….
1) 1 N/m
2) 0.01 N/m
3) 10N/m
4) 0.1 N/m
Answer:
4) 0.1 N/m
Current I = 0.1 A, Induction, B = 1T.
From F = BIl sinθ (where θ = 90°) ⇒ F = BIl
∴ Force per unit length \(\frac{\mathrm{F}}{l}\) = BI = 1 x 0.1 = 0.1N

Question 4.
A charged particle moves in a uniform magnetic field and its velocity makes an angle 6 with the direction of magnetic field. If 6 is other than 0°, 90° and 180°, its path is …
1) Helical path
2) Circle
3) Straight line
4) Elliptical path
Answer:
1) Helical path
At 0°, 180° the path is a straight line. At 90°, it is a circle.
At any other angle the path of the charged particle is helical path.

Question 5.
The magnetic field due to small length element \(\overrightarrow{\mathrm{dl}}\) of current carrying wire is given by \(\overrightarrow{\mathrm{dB}}\) = \(\frac{\mu_0 \mathbf{I}}{4 \pi} \frac{\overrightarrow{\mathbf{d l}} \times \overrightarrow{\mathbf{r}}}{\mathbf{r}^3},\), where \(\overrightarrow{\mathrm{r}}\) is the position of a point where \(\overrightarrow{\mathrm{dB}}\) is calculated. The direction of \(\overrightarrow{\mathrm{dl}}\) is
1) anti parallel to the direction electric current
2) parallel to \(\overrightarrow{\mathrm{r}}\)
3) perpendicular to \(\overrightarrow{\mathrm{r}}\).
4) parallel to the direction electric current
Answer:
4) parallel to the direction electric current
From Biot – Savart law, die length dl is always takes the direction of the current flow.

Question 6.
The magnetic field inside the very long solenoid is
1) non-uniform and varies along its radial direction
2) uniform and is parallel to its axis
3) non-uniform and varies along its length
4) uniform and is parallel to radial direction
Answer:
uniform and is parallel to its axis
For an ideal solenoid, the magnetic field lines inside are parallel, equally spaced and move along axis of the solenoid. This indicates the field is uniform, parallel to its axis.

Question 7.
Two straight long parallel wires, kept on an insulated horizontal table, carry electric currents in opposite directions. They
1) repel along horizontal direction
2) repel in upward direction
3) attract along horizontal direction
4) attract in vertically downward direction
Answer:
1) repel along horizontal direction
Parallel wires carrying current in opposite directions experience a repulsive force.

Question 8.
A current carrying coil is placed in a uniform magnetic field, it experiences in general
1) zero force and a constant torque
2) zero force and a variable torque
3) non-zero force and a constant torque
4) non-zero force and a variable torque
Answer:
2) zero force and a variable torque
In a uniform, magnetic field, the net force on a closed loop is always zero. Because the opposite forces cancel each other. If the current carrying coil is placed in magnetic field the coil rotates due to torque.
Torque 𝜏 = NIAB sinθ, So, Torque changes with rotation, (Angle θ varies)

Question 9.
The SI unit of magnetic moment is
1) ampere per meter
2) ampere per meter²
3) ampere – meter
4) ampere – meter²
Answer:
4) ampere – meter²
Magnetic moment M = NIA. Hoe N=number of turns, I = current in ampere, A = Area in m²
So, unit of M = ampere – metre²

Question 10.
A soft cylindrical iron is inserted into the coil of galvanometer because
1) it makes the magnetic field radial and increases the strength of the magnetic field
2) it makes the magnetic field radial and decreases the strength of the magnetic field
3) it makes die magnetic field uniform and increases the strength of the magnetic field
4) it makes the magnetic field uniform and decreases the strength of the magnetic field
Answer:
1) it makes the magnetic field radial and increases the strength of the magnetic field
The soft iron core has high permeability due to radial magnetic field strengths.

Question 11.
If the number of turns in the coil of galvanometer is increased by 4 times, then its voltage sensitivity
1) increases by 4 times
2) is unchanged
3) decreases by 4 times
4) increases by twice
Answer:
2) is unchanged
When no.of turns increases by 4 times, resistance also increases 4 times.
Thus if N = 4 N then we get R’ = 4R
Voltage sensitivity V_S = \(\frac{\mathrm{NBA}}{\mathrm{KR}}\) ⇒ V_S¹ = \(\frac{(4 N) B A}{K(4 R)}=\frac{N B A}{K R}\) = V_S
V_S is unchanged.

II. Fill in the Blanks

Question 1.
A neutron, moving through a magnetic field of 2T, has a speed of 10⁴ m/s at an angle of 30° with the direction of magnetic field. The magnitude of magnetic force experienced by it is _________
Answer:
zero.
The magnetic force is F = Bqv sinθ.
A neutron has a zero charge.
So, there is no field strength .Hence the magnetic force is zero.

Question 2.
A current carrying coil of area A is placed in a uniform magnetic filed B. The magnetic force experienced by the coil if it carries a current I is ________.
Answer:
zero.
For any closed coil, when it is situated in a uniform magnetic field, the vector sum of the magnetic forces on all segments of the loop in zero.

Question 3.
The distance, covered by a charged particle moving in a helical path, along the direction of magnetic field in one rotation is called ________.
Answer:
pitch.
The linear distance covered by a charged particle in one rotation in an helical path is known as the pitch.

Question 4.
The shape of the magnetic field lines due to a very lone straight wire is _________
Answer:
circular.
The magnetic field lines around a straight conductor form concentric circles centered on the wire.

Question 5.
The magnetic field inside the long solenoid carrying a current I is B = μ₀nI, where n is __________.
Answer:
number of turns per unit length.
In B = μ₀nI ‘n’ indicates number of turns per unit length.

Question 6.
Two parallel current carrying straight wires A and B exert force on each other. If the wire ‘A’ exerts a force 10 N due south on wire ‘B’, the force on wire A by wire B is _________
Answer:
10N due North.
According to Newton’s 3rd law, the force exerted by wire B on wire A is equal in magnitude & opposite in direction. Hence the direction is 10 N due North.

Question 7.
If one ampere current passes through each wire of two straight parallel wires of 1m separation, then the magnitude of magnetic force per unit length of each wire is _______.
Answer:
2 × 10^-7N/m
The force per unit-length between two parallel current carrying wires is \(\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2 \pi d}\)
\(\frac{\mathrm{F}}{\mathrm{~L}}=\frac{4 \pi \times 10^{-7} \times 1 \times 1}{2 \pi \times 1}\) = 2 × 10^-7N/m^-1

Question 8.
The resistance of an ideal ammeter is ________.
Answer:
zero.
An ideal Ammeter should have no resistance, as it doesn’t change the current.

Question 9.
The resistance of an ideal voltmeter is ________
Answer:
infinity.
An ideal Voltmeter should have infinite resistance, as it draws no current form the circuit.

Question 10.
The number of turns of a coil in a galvanometer is increased by 4 times, then its current sensitivity increases by ________ times.
Answer:
4
Current sensitivity I_s = \(\frac{\text { NBA }}{\mathrm{K}}\) ⇒ I_s ∝ N. So if N increases 4 times then I_s increases 4 times.

III. One Word Answer Questions

Question 1.
What is the relation between the units of magnetic field induction, Tesla and Gauss?
Answer:
Units of magnetic induction is Tesla. Relation: 1 Tesla(T) = 10⁴ Gauss(G)

Question 2.
The SI unit of magnetic field B is newton per meter times x. Then what is x?
Answer:
x = ampere. SI unit of magnetic induction is Newton per meter per ampere.

Question 3.
What is the path of a moving charged particle through a uniform magnetic field if it moves perpendicular to the direction of magnetic field?
Answer:
A charged particle moving perpendicular to a uniform magnetic field follows a circular path, because it continuously changes its direction while the speed is constant.

Question 4.
A point charge of strength q moves with linear momentum p in a circular path of radius r = p / K in a uniform magnetic field B . Express K in terms of q and B.
Answer:
K = qB
When a charge q moves in a circular path in a magnetic field, centripetal force = magnetic force
∴ \(\frac{m v^2}{r}\) = Bqv ⇒ r = \(\frac{m v}{B q}=\frac{p}{B q}\) [∵ p = mv] But given r = \(\frac{\mathrm{p}}{\mathrm{~K}}\) ……….. (2)
From (1) & (2) we get K = qB

Question 5.
What is the SI unit of permeability of free space?
Answer:
SI unit of μ₀ is henry per meter H/m. It is also equal to N/A²

Question 6.
The magnetic field due to small element of length \(\overrightarrow{\mathrm{dl}}\) of a current carrying wire is given by \(\overrightarrow{\mathrm{dB}}=\frac{\mu_0 \mathrm{I}[\overrightarrow{\mathrm{~d}} \mathrm{l} \times \overrightarrow{\mathrm{r}}]}{4 \pi} \frac{\mathrm{r}^{\mathrm{n}}}{\mathrm{r}^{\mathrm{n}}}\), where \(\overrightarrow{\mathrm{r}}\) is the position of a point where \(\overrightarrow{\mathrm{dB}}\) is calculated. What is the value of n?
Answer:
The vector from of Biot Savart’s law is \(\overrightarrow{\mathrm{dB}}=\frac{\mu_0 \mathrm{I}[\overrightarrow{\mathrm{~d}} \mathrm{l} \times \overrightarrow{\mathrm{r}}]}{4 \pi} \frac{\mathrm{r}^{\mathrm{n}}}{\mathrm{r}^{3}}\). So, the value of n = 3.

Question 7.
A circular wire, carrying an electric current, is placed in xy-plane and current flows in clockwise direction. What is the direction of magnetic field on its symmetry axis?
Answer:
The direction of magnetic field on its symmetry axis is along negative z-axis.

Question 8.
What is the magnetic moment of a circular coil of radius r, of number of turns N and of current I?
Answer:
Magnetic moment (M) of a circular coil M = NIA (or) M = NI(πr²)
Here N is the number of turns, r radius and A is the area of the coil.

Question 9.
What is the current sensitivity of galvanometer if its coil deflects by an angle Φ and current passes through it is I?
Answer:
Current sensitivity is S_I = \(\frac{\phi}{I}\)

Question 10.
What is the voltage sensitivity of galvanometer if its coil deflects by an angle and the voltage across it is V?
Answer:
Voltage sensitivity is S_V = \(\frac{\phi}{V}\)

IV. Very Short Answer Questions

Question 1.
What is the importance of Oersted’s experiment?
Answer:
Importance of Oersted’s Experiment :
Oersted’s experiment established the connection between electric current and magnetism. It showed that moving charge (current) produces magnetic field in its surrounding space.

Question 2.
What is the work done by the magnetic field on moving charge? Give reason.
Answer:
The work done by a magnetic field on a moving charge is zero.
Reason: Magnetic field \(\vec{F}\) is always perpendicular to the velocity \(\vec{v}\) of the charge.
Work done W = \(\vec{\mathrm{F}} \cdot \vec{\mathrm{ds}}=\int \vec{\mathrm{F}} \cdot \vec{\mathrm{v}}\) dt (dot product becomes zero)

Question 3.
What is the magnitude of magnetic force on a charged particle of strength q moving through a uniform magnetic field B at a velocity v which makes an angle 6 with magnetic field? When does this force become maximum?
Answer:
The magnitude of magnetic force (F) on a charge (q), moving with velocity (v) at an angle θ to a magnetic field (B) is given by F = Bqv sinθ.
This force becomes maximum when θ = 90°. Then sin0 = sin 90° = 1.
∴ F_max = Bqv

Question 4.
What is the magnitude of magnetic force 6n a current (i) carrying wire of length ‘l’ placed in a uniform magnetic field B if it makes an angle θ with magnetic field? When does this force become maximum?
Answer:
The magnitude of magnetic force (F) on a current (i) carrying wire of length (l) placed in a magnetic field (B) which makes an angle (θ), is given by F = Bil sinθ
This force becomes maximum when θ = 90°. Then sinθ = sin 90° = 1.
∴ F_max = Bil

Question 5.
Write expression for the magnetic field induction at a point on the axis of a circular current carrying coil. Obtain an expression for the magnetic field induction at its center.
Answer:
The magnetic induction on the axis of a circular coil at a distance x from its centre is B_x = \(\frac{\mu_0 \mathrm{NIR}^2}{2\left(x^2+R^2\right)^{3 / 2}}\)
Here N is number of the turns, I is the current passing through the coil,
R is radius of the coil, (XQ is permeability of free space:
Magnetic Induction at the centre of the circular coil:
Putting x = 0 in the above equation, we get magnetic induction the at centre of the coil B₀ = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}\)

Question 6.
Write the formula for the magnetic field due to infinite straight wire and explain the terms in it.
Answer:
Magnetic field due to infinite straight wire is B = \(\frac{\mu_0}{2 \pi} \frac{I}{r}\) . Here µ₀ is the permeability of free space, I is electric current, r is the perpendicular distance from line.

Question 7.
Express a relation among the quantities the speed of light in vacuum, the permeability of vacuum and the permittivity of vacuum.
Answer:
The speed of light in vacuum c = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\). Here µ₀ is permeability and ε₀ is the permittivity ofvaccum.

Question 8.
Two parallel conducting wires, separated by a distance d carry I₁ and I₂ currents. When do they attract and repel? What is the magnitude of force per unit length acting on any wire?
Answer:
The two parallel current conducting wires attract when die currents flow in the same direction. They repel when the currents flow in the opposite direction.
The magnitude of the force per unit length is f = \(\frac{\mathrm{F}}{l}=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{~d}}\)
Here µ₀ is permeability of free space and d is the distance between the wires.

Question 9.
What is the principle of moving coil galvanometer?
Answer:
Principle of MCG: The current carrying coil placed in a uniform magnetic field experiences a torque.

Question 10.
How do you convert a moving coil galvanometer into an ammeter?
Answer:
A galvanometer (G) can be converted into an ammeter (A) by connecting a very small shunt resistance (r_s) m parallel with it.

Question 11.
How do you convert a moving coil galvanometer into a voltmeter?
Answer:
A galvanometer (G) can be converted into a voltmeter (V) by connecting a high resistance . (R) in series with it.

V. Short Answer Questions

Question 1.
State and explain Biot-Savart law with a neat diagram.
Answer:
Biot-Savart Law:The magnitude of magnetic field induction due to small element of current carrying conductor is directly proportional to the strength of the current, length of the element, sine of the angle between position vector and the element; and inversely proportional to the square of the distance of the point from the element.

Explanation: Suppose RQ is a conductor carrying a current I. dl is a small length of element on the conductor.
The distance between dl and P is ‘r’.
Let θ be the angle between \(\overline{\mathrm{d} l}\) and radius vector \(\overline{\mathrm{r}}\)
The magnetic induction due to dl is dB
Here according to Biot-savart law
(i) dB ∝ I
(ii) dB ∝ dl
(iii) dB ∝ sin θ
(iv) dB ∝ \(\frac{1}{r^2}\)
Thus, dB ∝ \(\frac{\mathrm{Id} / \sin \theta}{\mathrm{r}^2}\) (or) dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l \sin \theta}{\mathrm{r}^2}\) Here \(\frac{\mu_0}{4 \pi}\) = 10^-7 Hm^-1

Question 2.
State Ampere’s circuital law. How it is used to calculate the magnetic field for a straight infinite current carrying wire with a neat diagram.
Answer:
Ampere’s Law: The line integral of the intensity of magnetic induction around a closed path (\(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{l}\)) is equal to μ₀ times the net current (i) enclosed by the path.
Thus \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{l}\) = μ₀. Here \(\mathrm{~d} \vec{l}\) = small element of the path, μ₀ = permeability of free space.

Expression for Magnetic field Induction:
Consider a long straight conductor carrying a current I.
Let P be a point at a distance r from the conductor.
Let r be the radius of the circle passing through the point P.
The circumference of the circle is 2πr.
Magnetic induction is same at all points on the circle. Consider a small element of length dl.
Now, \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{l}\) = \(\oint \) Bdl cos θ = B\(\oint \) dl cos θ
Angle between B and dl is zero ⇒ θ = 0 ⇒ cos θ = cos 0 = 1
∴ \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{l}\) = \(\vec{\mathrm{B}} \oint . \mathrm{d} \vec{l}\) = B(2πr) ………… (1)
From Ampere’s law, \(\oint \vec{\mathrm{B}} \cdot \mathrm{~d} \vec{l}\) = μ₀I …………… (2)
From (1) & (2) B(2πr) = μ₀I ⇒ B = \(\frac{\mu_0 I}{2 \pi r}\)

Question 3.
Derive an expression for the magnetic field induction at a point on the axis of a . current carrying circular coil using Biot-savart law with a neat diagram.
Answer:
Magnetic Induction on the axis a coil:
Consider a circular loop of radius R carrying a current I.

Let P be a point on the OX axis at a distance x from its centre O.
Consider a small element dl carrying the current I on the coil.
The distance of P from the element dl is r.
From the Biot-Savart law, the magnetic induction at P due to the element dl is dB = \(\frac{\mu_0}{4 \pi} \frac{||\mathrm{~d} \vec{l} \times \overrightarrow{\mathrm{r}}|}{\mathrm{r}^3}\)
The angle between dl and r is 90°. So \(\) = dlr sin90° = dlr
∴ dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l \mathrm{r}}{\mathrm{r}^3}=\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\mathrm{r}^2}\) ⇒ dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{x}^2+\mathrm{R}^2\right)}\) ………..(1) [From the figure, r² = x² + R²]
This dB is perpendicular to r. So its x-component dB_x = dB cos θ.
From the figure, cos θ = \(\frac{R}{r}=\frac{R}{\left(x^2+R^2\right)^{1 / 2}}\) ….(2)
From (1) & (2), dB_x – dB cos θ = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{x}^2+\mathrm{R}^2\right)} \frac{\mathrm{R}}{\left(\mathrm{x}^2+\mathrm{R}^2\right)^{1 / 2}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} / \mathrm{R}}{\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\)
Integrating on both sides and simplifying, we get B = Bx = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{IR} .2 \pi \mathrm{R}}{\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}=\frac{\mu_0 \mathrm{IR}^2}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\)
If the coil has N turns, magnetic filed induction B = \(\frac{\mu_0 \mathrm{NIR}^2}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\)

Question 4.
Derive an expression for the force acting on current carrying conductor placed in uni¬form magnetic field with a neat diagram.
Answer:
Force on a current carrying conductor in a magnetic field:
Consider a straight conductor of length 7’, cross sectional area ‘A’ and carrying a current ‘I’ placed in a uniform magnetic induction B as shown in figure.

Let us assume that the current flows through the conductor from left in X-Y plane and B makes an angle θ with the direction of current.
Force acting on a charge ‘q’ in the magnetic field is given by F¹ = qV_dB sin θ ………….. (1)
If ‘n’ represents number of free electrons per unit volume of the conductor, then
current in the conductor is given by I = nqV_dA ………….. (2)
Number of free electrons in the conductor, N = nlA …………. (3)
Total force acting on the entire conductor is,
F = F¹N = (qV_dB sinθ)(nLA) = (nqV_dA)(lBsinθ) F = BIlsin θ (∵ I = nqV_dA)
In vector form, above equation can be written as \(\overline{\mathrm{F}}=\mathrm{I}(\overline{l} \times \overline{\mathrm{B}})\)

Question 5.
Derive an expression for the force acting between two very long parallel current carrying conductors with a neat diagram.
Answer:
Force between two parallel conductors carrying Current:
Consider two straight parallel conductors a, b carrying currents I_a, I_b separated by a distance d .

Magnetic field produced by current-carrying conductor a is B_a = \(\frac{\mu_0 \mathrm{I}_{\mathrm{a}}}{2 \pi \mathrm{~d}}\) …………. (1)
Force acting on the conductor b by magnetic field B_a is F_ba = B_aI_bl ……….. (2)
From (1) & (2), we get F_ba = I_bl\(\left(\frac{\mu_0 \mathrm{I}_{\mathrm{a}}}{2 \pi \mathrm{~d}}\right)=\frac{\mu_0 \mathrm{I}_{\mathrm{a}} l}{2 \pi \mathrm{~d}}\) ………… (3)
The direction of the force Fba is towards conductor a.
Similarly, the force acting on the conductor a by magnetic field B is F_ab = \(\frac{\mu_0 \mathrm{I}_{\mathrm{a}} \mathrm{I}_{\mathrm{b}} l}{2 \pi \mathrm{~d}}\) ……… (4)
The direction of the force F_ab is towards conductor b.
From (3) & (4) we get F_ab = F_ba = \(\frac{\mu_0 \mathrm{I}_{\mathrm{a}} \mathrm{I}_{\mathrm{b}} l}{2 \pi \mathrm{~d}}\)

Question 6.
Explain bow the circular current carrying loop acts as magnetic dipole.
Answer:
When a current is passed through a circular current carrying loop, the current loop acts as a dipole with dipole moment \(\overrightarrow{\mathrm{M}}=\mathrm{IA} \hat{\mathrm{n}}\)
Consider a circualr coil of radius R carrying a current I.
Magnetic field at any point on the axial line of a circular current carrying coil at a distance x from its centre is B = \(\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{IR}^2}{\left(\mathrm{R}^2+\mathrm{x}^2\right)^{3 / 2}}\)
If R<2]
For a magnetic dipole, with dipole moment M, the magnetic induction at a point on its axis and far away from its centre is given by B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{x}^3}\) ………….. (2)
From (1) and (2) it follows that the current carrying loop behaves as a magnetic dipole with Magnetic moment M = IA .

Question 7.
DescriIe the method of conversion of galvanometer Into ammeter with a neat diagram. What is current sensitivity?
Answer:
A galvanometer (G) can be converted into an ammeter (A) by connecting a very small shunt resistance (r_s) parallel with it.

Derivation of Shunt Resistance
R_G = resistance of galvanometer, I_G = maximum current through galvanometer
r_S = shunt resistance, I total current to be measured
Then current through shunt: I_S = I – I_G
Since galvanometer and shunt are in parallel, potential difference across them is same:
I_GR_G = I_Sr_S
⇒ I_GR_G = (I – I_G)r_S
⇒ r_S = \(\frac{I_G R_G}{I-I_G}\)
This gives the value of shunt required to convert galvanometer into an ammeter.
Current sensitivity of the galvanometer is the deflection per unit current flowing through it.
Formula: Current sensitivity S_I = \(\frac{\theta}{I}=\frac{N A B}{K}\)

Question 8.
Describe the method of conversion of galvanometer into voltmeter, with a neat diagram. What is voltage sensitivity?
Answer:
A galvanometer (G) can be converted into a voltmeter (V) by connecting a high resistance (R) in series with it.

Derivation of Series Resistance
R_G = resistance of galvanometer
I_G = maximum current through galvanometer.
R = high resistance connected in series
V = maximum voltage to be measured
For full-scale deflection, current through galvanometer is I_G
Using Ohm’s 1gw:
V = I_G(R_G + R) ⇒ R_G + R = \(\frac{V}{I_G}\) ⇒ R = \(\frac{V}{I_G}\) – R_G
Voltage sensitivity of the galvanometer is the deflection obtained per unit voltage applied
Formula: Voltage sensitivity S_V = \(\frac{\theta}{V}=\left(\frac{N A B}{K}\right) \frac{1}{R}\)

VI. Long Answer Questions

Question 1.
Derive an expression for the force acting between two very long parallel current carrying conductors and hence define the ampere with a neat diagram?
Answer:
(a) Force between two parallel conductors carrying Current:
Consider two straight parallel conductors a, b carrying currents I_a, I_b separated by a distance d .

Magnetic field produced by current-carrying conductor a is B_a = \(\frac{\mu_0 \mathrm{I}_{\mathrm{a}}}{2 \pi \mathrm{~d}}\) …………. (1)
Force acting on the conductor b by magnetic field B_a is F_ba = B_aI_bl ……….. (2)
From (1) & (2), we get F_ba = I_bl\(\left(\frac{\mu_0 \mathrm{I}_{\mathrm{a}}}{2 \pi \mathrm{~d}}\right)=\frac{\mu_0 \mathrm{I}_{\mathrm{a}} l}{2 \pi \mathrm{~d}}\) ………… (3)
The direction of the force Fba is towards conductor a.
Similarly, the force acting on the conductor a by magnetic field B is F_ab = \(\frac{\mu_0 \mathrm{I}_{\mathrm{a}} \mathrm{I}_{\mathrm{b}} l}{2 \pi \mathrm{~d}}\) ……… (4)
The direction of the force F_ab is towards conductor b.
From (3) & (4) we get F_ab = F_ba = \(\frac{\mu_0 \mathrm{I}_{\mathrm{a}} \mathrm{I}_{\mathrm{b}} l}{2 \pi \mathrm{~d}}\)

(b) Ampere: When two infinitely long parallel conductors, carrying the same current are separated by a distance of 1m in vacuum, if the force per unit length on each conductor is 2 × 10^-7 Nm^-1 then the current flowing through each conductor is said to be one ampere.

Question 2.
Obtain an expression for the torque on a current carrying loop placed in a uniform magnetic field. Describe the construction and working of a moving coil galvanometer with a neat diagram.
Answer:
(a) Torque on a current carrying Loop in a Uniform Magnetic Field : Consider a rectangular current loop of length /, breadth b is suspended in a magnetic field B .
Let θ be the angle between the field and the normal to the loop.
The magnetic forces on the arms are given by F₁ = F₂ = ibB
The two forces constitute a couple.
The perpendicular distance between the forces is asin θ.
Torque = Force × Perpendicular distance
𝜏 = ibB × asinθ = iabBsinθ
= i A B sinθ [∵ Area of the loop ab = A]

(b) Construction and working of MCG:
Principle of MCG: The current-carrying coil placed in uniform magnetic field experience a torque.

Construction:
Moving coil galvanometer consists of a rectangular coil with a pivot arrangement in a strong radial magnetic field NS. The coil is free to rotate about a vertical axis. A soft iron cylinder is arranged at the centre of the coil to produce the magnetic field uniform. A spring is arranged to the coil to have a counter torque on die coil. A poiriter attached to the coil shows a reading on a semi-circular scale depending on the current flowing through it.

Working: If the current I to be measured is passed through the MCG, then the coil gets a deflection θ.
When a current I is passed through the coil suspended in a uniform radial magnetic field, it is subjected to two torques.
The deflecting torue acting on the coil is 𝜏 = BIAN …………. (1)
The restoring torque developed in the suspension is Cθ …………(2)
Here C is the torsional torque and 0 is the deflection of the coil
At equilibrium position, the deflecting magnetic torque = restoring torque due to spring
From (1) & (2), BIAN = Cθ
I = \(\frac{\mathrm{C} \theta}{\mathrm{BAN}}\). Put \(\frac{\mathrm{C}}{\mathrm{BAN}}\) = K
Hence, I = Kθ ⇒ I ∝ θ Here, K = \(\frac{\mathrm{C}}{\mathrm{BAN}}\) is called the constant of the galvanometer.
The relation I = Kθ explains the principle of MCG
By determining the values of K and θ, the current through MCG can be calculated.

Question 3.
Explain how can a Moving coil galvanometer be converted into
(a) Ammeter.
(b) Voltmeter with a neat diagram.
Answer:
A galvanometer (G) can be converted into an ammeter (A) by connecting a very small shunt resistance (r_s) parallel with it.

Derivation of Shunt Resistance
R_G = resistance of galvanometer, I_G = maximum current through galvanometer
r_S = shunt resistance, I total current to be measured
Then current through shunt: I_S = I – I_G
Since galvanometer and shunt are in parallel, potential difference across them is same:
I_GR_G = I_Sr_S
⇒ I_GR_G = (I – I_G)r_S
⇒ r_S = \(\frac{I_G R_G}{I-I_G}\)
This gives the value of shunt required to convert galvanometer into an ammeter.
Current sensitivity of the galvanometer is the deflection per unit current flowing through it.
Formula: Current sensitivity S_I = \(\frac{\theta}{I}=\frac{N A B}{K}\)

A galvanometer (G) can be converted into a voltmeter (V) by connecting a high resistance (R) in series with it.

Derivation of Series Resistance
R_G = resistance of galvanometer
I_G = maximum current through galvanometer.
R = high resistance connected in series
V = maximum voltage to be measured
For full-scale deflection, current through galvanometer is I_G
Using Ohm’s 1gw:
V = I_G(R_G + R) ⇒ R_G + R = \(\frac{V}{I_G}\) ⇒ R = \(\frac{V}{I_G}\) – R_G
Voltage sensitivity of the galvanometer is the deflection obtained per unit voltage applied
Formula: Voltage sensitivity S_V = \(\frac{\theta}{V}=\left(\frac{N A B}{K}\right) \frac{1}{R}\)

Question 4.
State and explain Biot-Savart law, thereby derive an expression for magnetic field induction at any point on the axis of circular current carrying coil with neat diagrams.
Answer:
Biot-Savart Law:The magnitude of magnetic field induction due to small element of current carrying conductor is directly proportional to the strength of the current, length of the element, sine of the angle between position vector and the element; and inversely proportional to the square of the distance of the point from the element.

Explanation: Suppose RQ is a conductor carrying a current I. dl is a small length of element on the conductor.
The distance between dl and P is ‘r’.
Let θ be the angle between \(\overline{\mathrm{d} l}\) and radius vector \(\overline{\mathrm{r}}\)
The magnetic induction due to dl is dB
Here according to Biot-savart law
(i) dB ∝ I
(ii) dB ∝ dl
(iii) dB ∝ sin θ
(iv) dB ∝ \(\frac{1}{r^2}\)
Thus, dB ∝ \(\frac{\mathrm{Id} / \sin \theta}{\mathrm{r}^2}\) (or) dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l \sin \theta}{\mathrm{r}^2}\) Here \(\frac{\mu_0}{4 \pi}\) = 10^-7 Hm^-1

Magnetic Induction on the axis a coil:
Consider a circular loop of radius R carrying a current I.

Let P be a point on the OX axis at a distance x from its centre O.
Consider a small element dl carrying the current I on the coil.
The distance of P from the element dl is r.
From the Biot-Savart law, the magnetic induction at P due to the element dl is dB = \(\frac{\mu_0}{4 \pi} \frac{||\mathrm{~d} \vec{l} \times \overrightarrow{\mathrm{r}}|}{\mathrm{r}^3}\)
The angle between dl and r is 90°. So \(\) = dlr sin90° = dlr
∴ dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l \mathrm{r}}{\mathrm{r}^3}=\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\mathrm{r}^2}\) ⇒ dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{x}^2+\mathrm{R}^2\right)}\) ………..(1) [From the figure, r² = x² + R²]
This dB is perpendicular to r. So its x-component dB_x = dB cos θ.
From the figure, cos θ = \(\frac{R}{r}=\frac{R}{\left(x^2+R^2\right)^{1 / 2}}\) ….(2)
From (1) & (2), dB_x – dB cos θ = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{x}^2+\mathrm{R}^2\right)} \frac{\mathrm{R}}{\left(\mathrm{x}^2+\mathrm{R}^2\right)^{1 / 2}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} / \mathrm{R}}{\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\)
Integrating on both sides and simplifying, we get B = Bx = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{IR} .2 \pi \mathrm{R}}{\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}=\frac{\mu_0 \mathrm{IR}^2}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\)
If the coil has N turns, magnetic filed induction B = \(\frac{\mu_0 \mathrm{NIR}^2}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\)

Textual Solved Problems

Question 1.
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field? .
Solution:
Weight of the wire = Upward magnetic force ⇒ m g = BIl
Here, m = 200 g = 0.2kg, g = 9.8m/s², I = 2A, l = 1.5m, B = ?
∴ 0.2 × 9.8 = 2 × 1.5 × B ⇒ B = 1.96/3 = 0.65 T

Question 2.
What is the radius of the path of an electron (mass 9 × 10^-31 kg and chaise 1.6 × 10^-19 C) moving at a speed of 3 × 10⁷ m/s in a magnetic field of 6 × 10^-4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10^-19 J).
Solution:
(i) Radius of the path r = \(\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{9 \times 10^{-31} \times 3 \times 10^7}{1.6 \times 10^{-19} \times 6 \times 10^{-4}}\) = 28 × 10^-2m = 28 cm

(ii) Frequency of electron v = \(\frac{v}{2 \pi r}=\frac{3 \times 10^7}{2 \times 3.14 \times 0.28}\) = 17 MHz

(iii) Energy E = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 9 × 10^-31 × (3 × 10⁷)² = 40.5 × 10^-17 J .
We know 1 eV = 1.6 × 10^-19 J ⇒ 1 J = 6.25 × 10¹⁸ ev = 6.25 × 10¹⁵ keV
∴ Energy in k eV is E = 4015 × 10^-17 × 6.25 × 10¹⁵ = 2.5 keV.

Question 3.
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil?
Solution:
Given N = 100, I = 1 A, R = 10 cm = 0.1 m, we know μ₀ = 4π × 10^-7 Hm^-1, B = ? .
We know Magnetic field B = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times 100 \times 1}{2 \times 0.1}\) = 6.28 × 10^-4 T

Question 4.
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid?
Solution:
Number of turns per unit length is, n = 500/0.5 = 10³ turns/m
Magnetic field inside the solenoid B = μ₀nI ⇒ B = 4π × 10^-7 × 10³ × 5 T B = 6.28 × 10^-3 T

Question 5.
The horizontal component of the earth’s magnetic field at a certain place is 3.0 × 10^-5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north?
Solution:
Given current I = 1 A, B = 3.0 × 10^-5 T
Force per unit length is f = IB sin θ
a) When current flows from east to west, θ = 90° ⇒ sin 90° = 1
⇒ f = 1 × 3 × 10^-5 = 3 × 10^-5 Nm^-1
b) When current flows from south to north θ = 0° ⇒ sin 0° = 0 ⇒ θ = 0° ⇒ f = 0

Question 6.
A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A.
a) What is the field at the centre of the coil ?
b) What is the magnetic moment of this coil ?
Solution:
Given N = 100, r = 10cm = 10 × 10^-2m = 0.1m, I = 3.2A
a) B_centre = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{r}}=\frac{2 \pi \times 10^{-7} \times 100 \times 3.2}{0.1}\) = 20.096 × 10^-4T = 2 × 10^-3T
b) Magnetic moment M = NIA = 100(3.2)π(0.1)² = 10.048 = 10 Am²

Question 7.
In the circuit, the current is to be measured. What is the value of the current if the ammeter shown
(a) is galvanometer with a resistance R_G = 60.000Ω
(b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance r_s = 0.02 Ω
(c) is an ideal ammeter with zero resistance?

Solution:
(a) Total resistance in the circuit is R_G + 3 = 63 Ω
∴ I = \(\frac{3}{63}\) = 0.048 A

(b) Resistarrce of the galvanometer converted to an ammeter = \(\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}_{\mathrm{s}}}=\frac{60 \times 0.02}{(60+0.02)}\) = 0.02Ω
Total resistance in the circuit 0.02Ω + 3Ω = 3.02Ω
∴ I = \(\frac{3}{3.02}\) = 0.99A
(c) For the ideal ammeter with zero resistance, I = \(\frac{3}{3}\) = 1A

Exercise Problems

Question 1.
What is the magnetic field at the center of a circular wire of radius 2 cm if it carries an electric current 1A?
Solution:
Given current I = 1 A; Radius R – 2cm – 0.02m, we know μ₀ = 4π × 10^-7
Magnetic induction B = \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times 1}{2 \times 0.02}\) = π × 10^-5 = 3.14 π × 10^-5 T

Question 2.
What is the magnetic field inside the solenoid of length 1mm and having 10⁴ turns and carrying a current 2A?
Solution:
Length of the solenoid l = 1 mm = 10^-3m, Number of turns N = 10⁴, Current I = 2A
Permeability μ₀ = 4π × 10^-7 Tm/A
Number of turns per unit length n =\(\frac{\mathrm{N}}{l}=\frac{10^4}{10^{-3}}\) = 10⁷
∴ Magnetic Induction B = μ₀nI = 4π × 10^-7 × 10⁷ × 2 = 8πT

Question 3.
A coil of area 10^-2m² and contains 10⁶ turns is used in a galvanometer. What is the maximum torque on it if the galvanometer carrying a current of 10^-4 A is placed in a magnetic field of 10² T?’
Solution:
Given area of the coil A = 10^-2m², Number of turns N = 10⁶, Current I = 10^-4 A
Magnetic induction B = 10²T, Torque will be maximum when θ = 90° ⇒ sin 90°= 1
Torque = BIAN sinθ
Maximum torque, 𝜏_max = BIAN = 10² ×10^-4 × 10^-2 × 10⁶ = 100 Nm

Question 4.
A Moving coil galvanometer, having a resistance of I ohm can measure a current 10 A. What Is the resistance of shunt required to measure a current 1A?
Solution:
Given galvanometer, resistance R_G = 1Ω, Current through galvanometer I_G = 10^-6 A,
Current to be measured, I = 1A, r_s = ?
ShuntResistance r_s = \(\frac{I_G R_G}{I-I_G}=\frac{10^{-6} \times 1}{1-10^{-6}}\) ≈ 10^-6Ω (∵ 1 – 10^-6 ≈ 1)

Question 5.
A current of 5A flows through each of two parallel long wires, the wires are 2.5 cms apart. Calculate the force acting per unit length of each wire.
Solution:
Current in each wire I₁ = I₂ = 5A
Distance between the wires r = 2.5 cm = 0.025 m. Also μ₀ = 4π × 10^-7
Force per unit length = \(\frac{\mathrm{F}}{l}=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{r}}=\frac{\left[4 \pi \times 10^{-7}\right][5][5]}{2 \pi(0.025)}\) = 2 × 10^-4 Nm^-1

Question 6.
A Moving coil galvanometer is placed in a radial magnetic field of 0.2T. The galvanometer coil has 200 turns and area of 1.6 × ^-4 m² the torsion constant at the suspension fibre is 10^-6 Nm/deg. Determine the maximum current that can be measured by this galvanometer, if its scale can accomodate a deflection of 45°.
Solution:
Magnetic induction B = 0.2T, Number of turns N = 200, Area of the coil A = 1.6 × ^-4 m²
Torsion constant K = 10^-6 Nm/deg, Maximum deflection 0 = 45°
For MCG at equilibrium, deflecting torque = restoring torque
⇒ BIAN = Kθ
Maximum current I_max = \(\frac{\mathrm{K} \theta}{\mathrm{BAN}}=\frac{10^{-6} \times 45^{\circ}}{0.02 \times 1,6 \times 10^{-4} \times 200}\) = 0.007 = 7 mA

Question 7.
A shunt of 6 Ω is connected across a galvanometer of resistance 29 Ω . Find the fraction of the total current passing through the galvanometer.
Solution:
Given Shunt resistance S = 6 Ω galvanometer resistance G = 294 Ω :
Fraction of total current \(\frac{I_G}{I}=\frac{S}{G+S}=\frac{6}{294+6}=\frac{6}{300}=\frac{1}{50}\)

Question 8.
A galvanometer with a coil of resistance 12 Ω shows full scale deflection for a current 2.5 Ω. How will you convert the galvanometer into (a) an ammeter of range 0 to 7.5A (b) a voltmeter of range 0 to 10V?
Solution:
Given resistance of Galvanometer G = 12Ω,
Full scale current deflection I_G = 2.5mA = 2.5 × 10^-3 A
a) Range of ammeter is from 0 to 7.5A ⇒ A= 7.5
Shunt S = \(\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}=\frac{2.5 \times 10^{-3} \times 12}{7.5-\left[2.5 \times 10^{-3}\right]}=\frac{0.03}{7.4975}\) = 0.004 Ω
To convert a galvanometer into a 7.5 ammeter, connect a 4 mΩ resistor in parallel.

b) Range of voltmeter is from 0 to 10V ⇒ V = 10
V = I_G(G + R) ⇒ R = \(\frac{V}{I_G}\) – G = \(\left[\frac{10}{2.5 \times 10^{-3}}\right]\) – 12 = 4000 – 12 = 3988 Ω
To convert a galvanometer into a 10V voltmeter, connect a 3988Ω resistor in series.

Objective Questions

Question 1.
The magnetic induction at the centre of a current carrying circular coil of radius 10cm is 5\(\sqrt{5}\) times the magnetic induction at a point on its axis, the distance of the point from the centre of the coil in cm is
1) 5
2) 10
3) 20
4) 25
Answer:
3) 20

Question 2.
A long straight wire carries an electric current of 2A. The magnetic induction at a perpendicular distance of 5m from the wire will be
1) 4 × 10^-8T
2) 8 × 10^-8T
3) 12 × 10^-8T
4) 16 × 10^-8T
Answer:
2) 8 × 10^-8T

Question 3.
Two parallel wires of length 9m each, are separated by a distance of 0.15m. If they carry equal currents in the same direction and exert a total force of 30 × 10^-7N on each other, the value of the current must be
1) 1.5 Amperes
2) 2.25 Amperes
3) 0.5 Ampere
4) 0.25 Amperes
Answer:
3) 0.5 Ampere

Question 4.
An electric current passes through a long straight wire. At a distance 5cm from the wire, the magnetic field is B. The field at 20cm from the wire would be
1) 2B
2) B/4
3) B/2
4) B
Answer:
2) B/4

Question 5.
In a galvanometer 5% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, the shunt resistance ‘S’ connected to the galvanometer is
1) 19G
2) \(\frac{\mathrm{G}}{19}\)
3) 20G
4) \(\frac{\mathrm{G}}{20}\)
Answer:
2) \(\frac{\mathrm{G}}{19}\)

Question 6.
A particle of mass 0.6gm and having charge of 25nC is moving horizontally with a uniform velocity 1.2xl04ra/s in a uniform magnetic field, then the value of the magnetic induction is
1) 0
2) 10T
3) 20T
4) 200T
Answer:
3) 20T

Question 7.
An electron remains undeflected when passing perpendicular to mutually perpendicular electric and magnetic fields. If the magnetic field is 8 Gauss and the electric field is 4000V/m, the velocity of the electron is
1) 2 × 10^-6m/s
2) 5.0 × 10^-6m/s
3) 6 × 10^-6m/s
4) 7.55 × 10^-6m/s
Answer:
2) 5.0 × 10^-6m/s

Question 8.
In the product
\(\vec{F}=q(\vec{v} \times \vec{B})=q \vec{v} \times\left(B \hat{i}+B \hat{j}+B_0 \hat{k}\right)\) For q = 1 and \(\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}\) and \(\vec{\mathrm{F}}=4 \hat{\mathrm{i}}-20 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\) What will be the complete expression for g ?
1) \(6 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}\)
2) \(-8 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\)
3) \(-6 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}\)
4) \(8 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\)
Answer:
3) \(-6 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}\)

Question 9.
A charge q moves in a region where electric field and magnetic field both exist, then force on it is
1) \(q(\vec{v} \times \vec{b})\)
2) \(q \vec{E}+q(\vec{v} \times \vec{B})\)
3) \(q \vec{E}+q(\vec{B} \times \vec{v})\)
4) \(q \vec{B}+q(\vec{E} \times \vec{v})\)
Answer:
2) \(q \vec{E}+q(\vec{v} \times \vec{B})\)

Question 10.
Tesla is the unit of
1) electric field
2) magnetic field
3) electric flux
4) magnetic flux
Answer:
2) magnetic field

Question 11.
An electron having mass m and kinetic energy E enter in uniform magnetic field B perpendicularly, then its frequency will be
1) eE/qvB
2) 2πm/eB
3) eB/2πm
4) 2m/eBE
Answer:
3) eB/2πm

Question 12.
A 10 eV electron is circulating in a plane at right angles to a uniform field at magnetic induction 10^-4 Wb/m² (= 1.0 gauss), the orbital radius of electron is 1) 11 cm
2) 18 cm
3) 12 cm
4) 16 cm
Answer:
1) 11 cm

Question 13.
A particle of mass m, charge Q and kinetic energy T enters in a transverse uniform magnetic field of induction \(\vec{\mathrm{B}}\). After 3 seconds the kinetic energy of the particle will be
1) T
2) 4T
3) 3T
4) 2T
Answer:
1) T

Question 14.
A beam, of electron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move
1) in a circular orbit
2) along a parabolic path
3) along a straight line
4) in an elliptical orbit
Answer:
1) in a circular orbit

Question 15.
A beam of electrons is moving constant velocity in a region having electric and magnetic fields of strength 20 Vm^-1 and 0.5T at right angles to the direction of motion of the electrons. What is the velocity of the electrons?
1) 8 ms^-1
2) 5.5 ms^-1
3) 20 ms^-1
4) 40 ms^-1
Answer:
4) 40 ms^-1

Question 16.
Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring is
1) \(\frac{\mu_0 q f}{2 \pi R}\)
2) \(\frac{\mu_0 q f}{2 R}\)
3) \(\frac{\mu_0 \mathrm{q}}{2 \mathrm{fR}}\)
4) \(\frac{\mu_0 \mathrm{q}}{2 \pi \mathrm{f}}\)
Answer:
2) \(\frac{\mu_0 q f}{2 R}\)

Question 17.
A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x- z plane. If the current in theloop is i. The resultant magnetic field due to the two semicircular parts at their common centre is
1) \(\frac{\mu_0 \mathrm{i}}{2 \sqrt{2} \mathrm{R}}\)
2) \(\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}}\)
3) \(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\)
4) \(\frac{\mu_0 \mathrm{i}}{\sqrt{2} \mathrm{R}}\)
Answer:
1) \(\frac{\mu_0 \mathrm{i}}{2 \sqrt{2} \mathrm{R}}\)

Question 18.
A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross¬section. The ratio of the magnetic fields B and B’, at radial distances a/2 and 2a respectively, from the axis of the wire is
1) 1
2) 4
3) 1/4
4) 1/2
Answer:
1) 1

Question 19.
The magnetic field at a distance r from a long wire carrying current i is 0.4 tesla. The magnetic field at a distance 2r is
1) 0.2 tesla
2) 0.8 tesla
3) 0.1 tesla
4) 1.6 tesla
Answer:
1) 0.2 tesla

Question 20.
A long solenoid 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of.the solenoid is (μ₀ = 4π × 10^-7 T m A^-1)
1) 6.28 × 10^-4T
2) 3.14 × 10^-4T
3) 6.28 × 10^-5T
4) 3.14 × 10^-5T
Answer:
1) 6.28 × 10^-4T

Question 21.
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is
1) B/2
2) B
3) 2B
4) 4B
Answer:
2) B

Question 22.
Two long parallel wires are at a distance of 1 metre. Both of them carry one ampere of current. The force of attraction per unit length between the two wires is
1) 5 × 10^-8N/m
2) 2 × 10^-8 N/m
3) 2 × 10^-7N/m
4) 10^-7 N/m
Answer:
3) 2 × 10^-7N/m

Question 23.
Two parallel wires in free space are 10cm apart and each carries a current of 10 A in the same direction. The force exerted by one wire on the other, per metre length is
1) 2 × 10^-4 N, repulsive
2) 2 × 10^-7N, repulsive
3) 2 × 10^-4 N, attractive
4) 2 × 10^-7N, attractive.
Answer:
3) 2 × 10^-4 N, attractive

Question 24.
If number of turns, area and current through a coil is given by n. A and i respectively then its magnetic moment will be
1) niA
2) n²iA
3) niA²
4) ni/\(\sqrt{\mathrm{A}}\)
Answer:
1) niA

Question 25.
A circular loop of area 0.01 m2 carrying a current 10 A, is held perpendicular to a magnetic field of intensity 0.1 T. The torque acting on the loop is
1) 0.001 Nm
2) 0.8 N m
3) zero
4) 0.01 N m.
Answer:
3) zero

Question 26.
A coil carrying electric current is placed in uniform magnetic field
1) torque is formed
2) e.m.f is induced
3) both (1) and (2) are correct
4) none of these
Answer:
1) torque is formed

Question 27.
In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be
1) 1/499 G
2) 499/500 G
3) 1/500 G
4) 500/499 G
Answer:
3) 1/500 G

Question 28.
A galvanometer of resistance, G, is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
1) \(\frac{\mathrm{G}}{(\mathrm{~S}+\mathrm{G})}\)
2) \(\frac{S^2}{(S+G)}\)
3) \(\frac{S G}{(S+G)}\)
4) \(\frac{G^2}{(S+G)}\)
Answer:
4) \(\frac{G^2}{(S+G)}\)

Question 29.
To convert a galvanometer into a voltmeter one should connect a
1) high resistance in series with galvanometer
2) low resistance in series with galvanometer
3) high resistance in parallel with galvanometer
4) low resistance in parallel with galvanometer.
Answer:
1) high resistance in series with galvanometer

Question 30.
A galvanometer having a resistance of 9 ohm is shunted by a wire of resistance 2 ohm. If the total current is 1 amp, the part of it passing through the shunt will be
1) 0.2 amp
2) 0.8 amp
3) 0.25 amp
4) 0.5 amp
Answer:
2) 0.8 amp

Question 31.
To convert a galvanometer into a ammeter, one needs to connect a
1) low resistance in parallel
2) high resistance in parallel
3) low resistance in series
4) high resistance in series.
Answer:
1) low resistance in parallel

Regular practice with AP Inter 2nd Year Physics Study Material Chapter 14 Semiconductor Devices Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Physics 14th Lesson Semiconductor Devices Questions and Answers

I. Multiple Choice Questions

Question 1.
Following is an example of a compound semiconductor [ ]
1) Al₂O₃
2) SiO₂
3) H₂O
4) GaAs
Answer:
4) GaAs
Gallium Arsenide (Ga As) is classified as a compound semiconductor

Question 2.
Which of the following is used as pentavalent impurities for doping process?[ ]
1) Arsenic
2) Boron
3) Indium
4) Aluminium
Answer:
1) Arsenic
Arsenic is used as a Pentavalent depend because, it contributes a base electron to the semi conductor lattice, creating an n-type material with high increased electrical conductivity.

Question 3.
For an n-type semiconductor n_e = number of conduction electrons and n_h, = number of holes, then ____. [ ]
1) n_e = n_h
2) n_e > n_h
3) n_e < n_h
4) n_e = (n_h/2)
Answer:
2) n_e > n_h
For n-type semiconductor, n_e > n_r, because the material is intentionally doped with pentavalent impurities.

Question 4.
The energy band gaps of silicon and germanium are [ ]
1) 0.7eV, 1.1eV
2) 1eV, 0.7eV
3) 1.1eV,0.7eV
4) 2eV, 1.5eV
Answer:
3) 1.1eV,0.7eV
Silicon and Germanium have different atomic sizes, nuclear charges and crystal binding forces. Because of this it takes 1.1 ev of energy the break a covalent bond in Silicon, compared to only 0.7ev in Germanium.

Question 5.
On the basis of electrical conductivity, which one of the following material has the smallest resistivity? [ ]
1) Germanium
2) Silicon
3) Glass
4) Silver
Answer:
4) Silver
Silver has lowest resistivity 1.59 × 10^-8Ω. It is due to its unique atomic structure how electronegativity, high concentration of free electrons.

Question 6.
In case of full wave rectification, if the input frequency is 60 Hz, then the output frequency would be [ ]
1) 60 Hz
2) 120 Hz
3) 100 Hz
4) 150 Hz
Answer:
2) 120 Hz
In a full wave rectifier the input frequency n = 60 Hz then its output frequency is 2n = 120 Hz

Question 7.
The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them. [ ]
1) No current will flow in p-type, current will only flow in n-type
2) Current in n-type = Current in p-type
3) Current in p-type > Current in n-type
4) Current in n-type > Current in p-type
Answer:
4) Current in n-type > Current in p-type
I_n > I_p. At equal charge carrier concentration, current in entire (mobility of electrons) is higher than current in p-type (mobility of holes)

Question 8.
For a p-n junction diode, the current in reverse bias is in the order of [ ]
1) Few mA
2) Few amperes
3) Few µA
4) Few kilo amperes
Answer:
3) Few µA
In a p-n junction diode, the reverse bias current is extremely small in the order of µA for Ge

Question 9.
The increase in the width of the depletion region in a p-n junction diode is due to [ ]
1) reverse bias only
2) both forward bias and reverse bias
3) high doping concentrations
4) forward bias only
Answer:
1) reverse bias only
The increase in the width of the depletion region in a p-n junction diode is due to reverse bias only

Question 10.
A Ge specimen is doped with aluminium. The concentration of acceptor atoms is 10²¹ atoms/m³. Given that the intrinsic concentration of electrons in the specimen is 10¹⁹/m³. The new electron concentration is [ ]
1) 10¹⁷/m³
2) 10¹⁸m³
3) 10⁴/m³
4) 10²/m³
Answer:
1) 10¹⁷/m³
When concentration of electron is n, hole is p and inter intrinsic carrier is n; then nP = \(\mathrm{n}_{\mathrm{i}}^2\)
∴ n = \(\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{P}}\) = \(\frac{\left(10^{19}\right)^2}{10^2}\) = \($\frac{10^{38}}{10^{21}}$\) = 1017m³

Question 11.
Which of the following circuits represents a forward biased diode? [ ]
Choose the correct answer from the options given below:

1) (C) and (E) only
2) (A) and (D) only
3) (B), (D) and (E) only
4) (B), (C) and (E) only
Answer:
4) (B), (C) and (E) only
A diode in forward biased, V_anode > V_cathode. Among the given (B)(C) (E) are correct.

II. Fill in the Blanks

Question 1.
In a vacuum tube, the electrons are supplied by a ___.
Answer:
cathode
In a vacuum tube, the electrons are supplied by a cathode.

Question 2.
In vacuum tubes, electrons flow from the cathode to the anode in ___ direction.
Answer:
only one
In vacuum tubes, electrons flow from the cathode to the anode in only one direction.

Question 3.
The conduction in a semiconductor possible due to motion of ___ in the conduction band and ___ in the valence band.
Answer:
The conduction in a semiconductor possible due to motion of electrons in the conduction band and holes in the valence band.

Question 4.
The gap between the top of the valence band. and bottom of the conduction band is called the ____.
Answer:
energy band gap/forbidden energy gap.
The gap between the top of the valence band and bottom of the conduction band is called the energy band gap/forbidden energy gap.

Question 5.
The semiconductor elements Ge and Si crystal structure is almost similar to ____ structure.
Answer:
diamond cubic crystal
The semiconductor elements Ge and Si crystal structure is almost similar to diamond cubic crystal structure.

Question 6.
In intrinsic semiconductors, the number of free electrons(n_e) = number of holes (n_h) = n_i where n_i is called ____.
Answer:
intrinsic carrier concentration
In intrinsic semiconductors, the number of free electrons(n_e) = number of holes (n_h) = n_i where n_i is called intrinsic carrier concentration.

Question 7.
The pentavalent dopant donates one extra electron for conduction and hence this impurity is called ___.
Answer:
donor
The pentavalent dopant donates one extra electron for conduction and hence this impurity is called donor.

Question 8.
In an n-type Si semiconductor, the donor energy level E_D is slightly ___ the bottom of the conduction band (E_C).
Answer:
below
In an n-type Si semiconductor, the donor energy level E_D is slightly below the bottom of the conduction band (E_C).

Question 9.
At Thermal equilibrium the electron and hole Concentration in an intrinsic semiconductor is ___.
Answer:
equal.
At Thermal equilibrium the electron and hole Concentration in an intrinsic semiconductor is equal.

Question 10.
In the p-n juñction diode symbol, the direction of arrow indicates ____.
Answer:
direction of conventional current
In the p-n junction diode symbol, the direction of arrow indicates direction of conventional current.

III. One Word Answer Questions

Question 1.
At what temperature the intrinsic semiconductor will behave like an insulator?
Answer:
At absolute zero temperature (0 K) intrinsic semi conductor behaves like an insulator

Question 2.
What is the ratio of number of holes and the number of conduction electrons in an intrinsic semiconductor?
Answer:
The ratio of the number of holes (n_h) to the number of conduction electrons (n_e) in an intrinsic semiconductor is 1:1.

Question 3.
What is the overall charge of the p-type semiconductor crystal?
Answer:
The overall charge of p-type semi conductor crystal is electrically neutral (zero).

Question 4.
Draw the circuit symbol of the p-n junction diode.
Answer:
Circuit symbol of the p-n junction diode:

Question 5.
For a p-n junction diode, built in potential = V₀ and applied voltage is V (V >V₀) and diode is in forward bias then what is the height of effective barrier?
Answer:
The effective barrier height = V₀ – V.
The effective barrier height (potential) of a forward biased diode is reduced by the applied voltage (V) from initial voltage (V₀).

Question 6.
What is the forbidden energy gap in Ge semiconductor crystal in terms of Joules?
Answer:
Forbidden energy gap in Ge at room temperature Eg = 0.72 × 1.6 × 10^-19 J = 1.15 × 10^-19 J

Question 7.
What is the ngme of the current due to motion of charge carriers when external electric field is applied.
Answer:
The current produced by the motion of charge carries under the influence of an external electric field is known as drift current.

Question 8.
At thermal equilibrium condition what will be the product of number of electrons and number of holes in any semiconductor material?
Answer:
At thermal equilibrium, the product of the concentration of electrons (n_e) and the concentration of holes (n_h) in any semi conductor material is given by n_en_h = \(\mathbf{n}_{\mathbf{i}}^2\).
Here n_i is the intrinsic carrier concentration.
This is known as the “Law of mass Action”.

Question 9.
What is the energy band which includes the energy levels of the valence electrons?
Answer:
The valence band is the energy band includes the energy levels of the valence electrons. It is the highest occupied energy band in a material.

Question 10.
What is the name of the process “adding impurities to the intrinsic semiconductor to make extrinsic semiconductor”?
Answer:
Doping is the process of adding impurities to an intrinsic semiconductor to make extrinsic semi conductor.

Question 11.
The output voltage in the following circuit is (Consider diodes are ideal)
Answer:

Output voltage is equal to zero.
Reason: The diode D₂ conducts and acts as a short circuit to ground.

Question 12.
What is the output frequency of a half-wave rectifier if the input voltage frequency is 50 Hz?
Answer:
If the input is 50Hz, out is also 50Hz.
Reason: The output frequency of a rectifier is equal to input voltage frequency.

Question 13.
What is the current through an ideal p-n junction diode shown in the following figure.

Answer:
Current through the diode is i = 0 A.

Reason: For an ideal P-N junction diode in reverse bias, the resistance is considered infinite and hence no current flows through it.

Question 14.
A semiconductor is known to have an electron concentration of 8 × 10¹³ /cm³ and a hole concentration 5 × 10/cm³. What type of the semiconductor is it?
Answer:
It is an n-type semiconductor because n_e > n_h.
Electron concentration ne = 8 × 10¹³ /cm³ ; Hole concentration n_h = 5 × 10² /cm³. Here the electron concentration is much higher than the hole concentration.

IV. Very Short Answer Questions

Question 1.
What are intrinsic and extrinsic semiconductors ?
Answer:
Intrinsic Semiconductor:
Pure semiconductor like silicon is called intrinsic semiconductor.
In an intrinsic semiconductor, number of electrons(n_e) = number of holes(n_h).
Extrinsic Semiconductor : A semiconductor doped with trivalent or pentavalent impurity is called extrinsic semiconductor.

Extrinsic semiconductors are of two types,

1. n-type semiconductor and
2. p-type semiconductor.

Question 2.
What is an n-type semiconductor?
What are the majority and minority charge carriers in it?
Answer:
n-type Semiconductor:
Silicon or Germanium doped with pentavalent impurity like Arsenic is called n-type semiconductor. The pentavalent atoms add extra electrons and make it n-type semiconductor.
In n-type semiconductor, majority charge carriers are electrons and minority charge carriers are holes.

Question 3.
What is a p-type semiconductor ? What are the majority and minority charge carriers in it?
Answer:
p-type Semiconductor:
Silicon or Germanium doped with trivalent impurity like Indium is called p-type semiconductor. The trivalent atoms add extra holes and make it p-type semiconductor.
In p-type semiconductor, majority charge carriers are holes and minority charge carriers are electrons.

Question 4.
How is a battery connected to a junction diode in
(1) forward and
(2) reverse bias?
Diode in Forward Bias
Answer:
1) Forward bias:

In forward bias, a battery is connected to the diode such that its positive terminal is towards p-end of diode.

2) Reverse bias:
In reverse bias, a battery is connected to the diode such that its positive terminal is towards n-end of diode.

Question 5.
What do you mean by donor impurity and acceptor impurity?
Answer:
Donor impurities are atoms with five valence electrons-like Phosphorus (Pentavalent) added to semiconductors to donate free electrons forming n-type materials.
Acceptor impurities are atoms with three valence electrons- like Boron (trivalent) added to semiconductors to create vacancies (holes) forming p-type materials.

Question 6.
What is reverse saturation current of p-n junction diode?
Answer:
Reverse saturation current is the small amount of current that flows through a p-n junction diode when it is reverse biased. It is caused by the flow of minority carriers.

Question 7.
Define depletion layer and what is the breakdown voltage of p-n junction diode.
Answer:
Depletion layer: It is the narrow region on either side of the p-n junction which is free from mobile charge carriers.
Breakdown voltage: It is the voltage in a reverse biased p-n junction at which the current suddenly rises to a large values.

Question 8.
What happens to the width of the depletion layer in a p-n junction diode when it is

1. forward-biased and
2. reverse biased?

Answer:

1. In forward bias, the width of the depletion layer of a diode decreases
2. In reverse bias, the width of the depletion layer of a diode increases.

Question 9.
What is the current through the battery in the circuit as shown in figure, here diodes are ideal?

Answer:
Both the diodes are in forward bias and both are ideal.

Question 10.
What is rectification? Which device is used as rectifier?
Answer:
Rectification is the process of converting an alternating current in to direct current.
The device used for this rectification process is called a rectifier.

V. Short Answer Questions

Question 1.
Describe energy band diagram of intrinsic semiconductors at 0 K.
Answer:
Intrinsic semiconductors:
Intrinsic semiconductors are made up of pure form of semi conductor. These atoms are tetravalent.
They behave like insulators at OK.

At 0K, the valence band is completely filled with electrons and conduction band is empty.
Thus there are no free electrons to conduct electricity. Hence they act as insulators at 0K.

At 0K there is a small energy gap between the top of the valence band and the bottom of the conduction band. Since, no thermal energy is available at 0 K, no electrons can jump across this gap.

At higher temperatures (T > 0K) thermal energy excites some electrons from valence band to the conduction band and leaving equal number of holes there.

Question 2.
How conductors, insulators and semiconductors are classified based on the energy band diagram?
Answer:
1) Conductors: In conductors, the conduction band and the valence band overlap or are very close to each other. Conduction band is partially filled, and there is no forbidden energy gap (band gap) between the valence band and conduction band. This allows electrons to move freely, resulting in high electrical conductivity.

2) Insulators: In insulators, the valence band is completely filled, and the conduction band is empty. There is a large forbidden energy gap (E_g > 3eV) between the valence band and conduction band. This large gap prevents electrons from jumping to the conduction band resulting in very low electrical conductivity.

3) Semiconductors: In semiconductors, the valence band is also filled, but the conduction band is either empty or partially filled. The forbidden energy gap is very small (E_g < 3eV). This smaller gap allows some electrons to jump into the conduction band due to thermal excitation, resulting in moderate electrical conductivity.

Question 3.
What are n-type and p-type semiconductors? Explain the process of formation of p-n junction diode.
Answer:
n-type Semiconductor. Silicon or germanium doped with pentavalent impurity like arsenic is called n-type semiconductor. The pentavalent atoms add extra electrons and make it n-type semiconductor. In n-type semiconductor, majority charge carriers are electrons and minority charge carriers are holes.

p-type Semiconductor: Silicon or germanium doped with trivalent impurity like indium is called p-type semiconductor. The trivalent atoms add extra hole’s and make it p-type semiconductor. In p-type semiconductor, majority charge carriers are holes and minority charge carriers are electrons.

Process of formation of p-n junction diode:

When a semiconductor material such as silicon or Germanium crystal is doped such that one side of it is p-type and the other side is n-type then the interface is called p-n junction and the device is called p-n junction diode.

When p-n junction is formed, the free electrons which are in higher concentration on n-side diffuse over to p-side and combine with holes and become neutral. Similarly, hole’s which are in higher concentration on p-side diffuse over to n-side and combine with electrons and become neutral.

This results in the formation of a narrow region on either side of the junction which is free from mobile charge carriers. This region is called depletion layer. This layer acts as a potential barrier between p-type and n-type material of diode.

Question 4.
Draw and explain the current-voltage (I -V) characteristic curves of a junction diode in forward and reverse bias.
Answer:
Voltage – Current Characteristics of Diode:
A diode is connected in forward bias and readings are noted with applied potential differences (PDs) and obtained currents.’
The same process should be repeated in reverse bias also.

With the measured values a graph is drawn by taking voltage on x-axis and current on y-axis. The graph obtained is called characteristic (I-V) curve of the diode.

Forward bias – Threshold voltage (V₀)
Threshold voltage is the voltage in a forward bias biased p-n junction at which the current suddenly rises to a large values.
The curve shows that the diode Conducts electric current during the forward bias but can not conduct during the reverse bias. Due to this property, diode is used as a rectifier.

Reverse bias – Breakdown voltage (V_B)
Breakdown voltage: It is the voltage in a reverse biased p-n junction at which the current suddenly rises to a larger values.

Question 5.
Describe how a semiconductor diode is used as a half wave rectifier.
Answer:
Rectification: The process of converting alternating current (AC) into direct current (DC) is called rectification.
Half Wave Rectifier: A circuit which rectifies half of the AC wave is called Half wave rectifier.

Diode as a Halfwave Rectifier: A single diode is connected in series to a load resistance R_L in the secondary circuit of a transformer.

Working:

• During the positive half cycle, the diode is forward biased and the current flows through the diode.
• During the negative half cycle, the diode is reverse biased and the current does not flow through the diode.
• Thus, half wave is rectified using a single diode.
• Efficiency of Halfwave rectifier η = \(\frac{0.406 R_L}{r_f+R_L}\)
• Maximum Efficiency of Halfwave rectifier is 40.6%

Question 6.
What is rectification? Explain the working of a full wave rectifier.
Answer:
Rectification: The process of converting alternating current (AC) into direct current (DC) is called rectification.

Full Wave Rectifier: The circuit which rectifies both half cycles of an AC wave is called Full Wave Rectifier.

Working:

• During the first half cycle, the voltage on positive. It acts as forward bias. Hence current passes through R_L. During this cycle, D₂ is reverse biased. Hence no current passes through D₂.
• During šecond half cycle, the D₂ is forward biased and D₁ is reverse biased. Hence, current passes through R_L.
• Thus the full Wave is rectified using two diodes.
• Efficiency of full wave rectifier is η = \(\frac{0.812 R_L}{r_f+R_L}\)
  Here r_f = diode forward resistance, R_L = load resistance.
• Maximum Efficiency of full-wave rectifier is 81.2%

VI. Long Answer Questions

Question 1.
What is a junction diode ? Explain the formation of depletion region at the junction. Explain the variation of depiction region in forward and reverse-biased conditions.
Answer:
p-n junction Diode.The device having a p-n junction is called p-n junction diode.
The region where p-type and n-type semiconductors in contact is called p-n junction.

The symbol of p-n junction diode:

Depletion Layer: In a p-n junction diode, the electrons of n-side and holes of p-side diffuse to opposite sides and form a depletion layer at p-n junction. It is called so because, the free charge carriers are depleted from the region. The thickness of depletion layer is about one tenth of a micrometer.

Forward Bias: If a battery is connected to junction diode such that its positive terminal is towards p-end and its negative terminal is towards n-end, the diode is said to be in forward bias.

In forward bias, the applied voltage overcomes the barrier potential because the applied potential is greater and opposite to barrier potential. As a result, the width of the depletion layer decreases. The charge carriets can easily cross the depletion layer and a current passes through the diode.

Reverse Bias: If a battery is connected to a junction diode such that its positive terminal is towards n-end and its negative terminal is towards p-end, the diode is said to be in reverse bias.

In reverse bias, the applied voltage cannot overcome the barrier potential because the applied potential is along the barrier potential. As a result, the width of the depletion layer increases. The charge carriers can not cross the depletion layer and no current passes through the diode.

Question 2.
What is a rectifier ? Explain the working of half wave and full wave rectifiers with diagrams.
Answer:
Rectifier: The device that converts AC into DC is called rectifier.
Rectification: The phenomenon of converting AC into DC is called rectification.

A) Half Wave Rectifier:
The half wave rectifier circuit is as shown in the figure. A single diode is connected in series to a load resistance R_L in the secondary circuit of a transformer.

• As the current is AC, the direction of the current in the secondary winding changes after every half cycle.
• During the positive half cycle, the diode is forward biased and the current flows through the diode.
• During the negative half cycle, the diode is reverse biased and the current does not flow through the diode.
• The current through the load resistance is discontinuous pulsating DC as shown in the figure.
• The discontinuous pulsating DC can be made pure DC by using a filter circuit.

B) Full Wave Rectifier: The full wave rectifier circuit consists of two diodes (D₁, D₂), a load resistance (R_L) in the secondary circuit of a centre tapped transformer as shown in the figure.

• As the current is AC, the direction of the current in the secondary winding of the transformer changes after every half cycle with respect to its centre tap C.
• During the first half cycle, the voltage on D₁ is positive. It acts as forward bias to D₁. Hence current passes to R_L through D₁. During this cycle, D₂ is reverse biased. Hence no current passes through D₂.
• During second half cycle, the D₂ is forward biased and D₁ is reverse biased. Hence, current passes to R_L through D₂.
• Thus the full wave is rectified.
• The output of full wave rectifier contains small variations as shown in the figure below. It can be made pure DC by using a filter circuit.

Exercise Problems

Question 1.
In a pure Intrinsic semiconductor, the number of conduction electron is 6 × 10¹⁹ per cubic meter. How many holes are there in a sample of size 1 cm × 2 cm × 2mm?
Answer:
Intrinsic concentration n_i = 6 × 10¹⁹, m^-3;
Volume v = 1cm × 2cm × 0.2cm = 0.4cm³ = 0.4 × 10^-6m³
Total number of holes = n_i × v = 6 × 10¹⁹ × 0.4 × 10^-6 = 2.4 × 10¹³

Question 2.
Pure silicon at 300K has an equal number of electrons (n_i) and the hole (n_h) concentrations 5 × 10¹⁶ m^-3. When it was doped with the indium then the hole concentration (n_h) increases to 4.5 ×10²² m^-3 then find the concentration of electrons after doping.
Answer:
Given n_i = 5 × 10¹⁶m^-3; n_h = 4.5 × 10²²m^-3
= 5.55 × 10¹⁰m³

Question 3.
An AC voltage with peak value 20 V is connected in series a diode and load resistance of 500Ω exists across the diode. Calculate the peak current through the diode and peak voltage across the load resistance when the diode is ideal.
Answer:
Given V_peak = 20V, Resistance R = 500Ω
∴ I_peak = \(\frac{V_{\text {peak }}}{R}\) = \(\frac{20}{500}\) = 0.04 = 40mA

Question 4.
A sample of pure silicon is doped with a trivalent impurity, where each impurity atom creates a hole. The concentration of holes in pure silicon is 7 × 10¹⁵ holes per cubic metre, and the density of silicon is 5 × 10²⁸ atoms per cubic metre. The doping increases the hole conceñtration by a factor of 120. Calculate approximately the proportion in which the impurity is added.
Answer:
Given Initial concentration n_i = 7 × 10¹⁵ m^-3, Silicon density n_Si= 5 × 10²⁸ atoms/m³
Doping increase n_h by a factor of 120
New hole concentration after doping n_h = factor × n_i = 120 × (7 × 10¹⁵) = 8.4 × 10¹⁷ holes / m³
No.of impurity atoms n_im = 8.4 × 10¹⁷ atoms / m³
Proportion of impurity = \(\frac{\mathbf{n}_{\mathbf{S i}}}{\mathbf{n}_{\mathbf{i m}}}\) = \(\frac{5 \times 10^{28}}{8.4 \times 10^{17}}\) = 5.98 × 10¹⁰
Proportion of impurity = one in 5.98 × 10¹⁰ silicon atoms.

Question 5.
A potential barrier of 0.8V exists across the p-n junction. If the depletion region is 800 nm wide, then what is the intensity of electric field with in this region is?
Answer:
Barrier potential V = 0.8V
Width of depletion region d = 800 nm = 800 × 10^-9 m = 8 × 10^-7 m
∴ Electric field Intentisty E = \(\frac{V}{d}\) = \(\frac{0.8}{8 \times 10^{-7}}\) = 10⁶ V/m

Question 6.
The diode shown in the circuit is a silicon diode, then find the potential difference between the points A and B ?

Answer:
Battery voltage V = 6V
Here, the diode is reverse bias. It acts as an open switch. No current flows through the circuit.
∴ The potential at A is V_A = 6V. Also V_B = 0
∴ The potential difference V_AB = V_A – V_B = 6 – 0 = 6V

Question 7.
A diode having potential difference 0.7V across its junction which does not depend on current, is connected in series with resistance 100 ohms across source, if 0.1A passes through resistance. Then find the voltage of the source?
Answer:
Voltage drop across the diode V_d = 0.7V, Resistance R = 100Ω
Current in the circuit I = 0.1 A Voltage across R is VR = I × R = 0.1 × 100 = 10V
Source voltage V_s = V_d + V_R = 0.7 + 10 = 10.7 V

Question 8.
A 2V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in one case battery’s positive terminal is connected to A and in another case when the positive terminal of the battery is connected to the B will respectively be?

Answer:
Terminal A : Diode D₁ is forward biased. So, current flows through 5Ω resistor
∴ current I₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\) = \(\frac{2}{5}\) = 0.4A
Terminal B :Diode D₂ is forward biased So, current flows through 10Ω resistor
∴ current I₂ = \(\frac{\mathrm{V}}{\mathrm{R}_2}\) = \(\frac{2}{10}\) = 0.2A

Question 9.
The I-V characteristics of a p-n junction diode in forward bias are shown in the following figure. Find the ratio of dynamic resistance corresponding to forward bias voltages of 2V and 4V?

Answer:
(i) Dynamic resistance at 2V voltage
Voltage at V₁ = 2V; current I₁ = 5mA
Voltage at V₂ = 2.1V; current I₂ = 10mA
∴ ΔV = V₂ -V₂ = 2.1 – 2 = 0.1 V
ΔI = I₂ – I₁ = 10 – 5 = 5 × 10^-3A
Resistance Rd₁ = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}\) = \(\frac{0.1}{5 \times 10^{-3}}\) = 20Ω

(ii) Dynamic resistance at 4V voltage.
Voltage at V₃ = 4V, current I₃ = 200mA
Voltage at V₄ = 4.2V, current I₄ = 250mA
ΔV = V₄ – V₃ = 4.2 – 4 = 0.2V
ΔI = I₄ – I₃ = 250 – 200 = 50 = 50 × 10^-3A
Resistance R d₂ = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}\) = \(\frac{0.2}{50 \times 10^{-3}}\) = \(\frac{200}{50}\) = 4Ω
∴ Ratio = \(\frac{\mathrm{Rd}_1}{\mathrm{Rd}_2}\) = \(\frac{20}{4}\) = 5 : 1

Question 10.
Two ideal diodes are connected in the network as shown ¡n the figure. Then find the equivalent resistance
between A and B?

Answer:

20Ω resistors are in parallel,
∴ \(\frac{1}{R_P}\) = \(\frac{1}{20}\) + \(\frac{1}{20}\) = \(\frac{2}{20}\) = \(\frac{1}{10}\) ⇒ R_P = 10Ω
Equivalent resistance R_eq = R_P + 15Ω = 10 + 15 = 25Ω

Question 11.
In the given circuit the diodes are IDEAL, then find the total resistance of circuit and the current reading in the ammeter.

In parallel combination, \(\frac{1}{R_P}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) ⇒ R_p = 2Ω
R_total = R_p – 4 = 2 + 4 = 6 ;Voltage V = 6V and R_total = 6Ω
∴ Current I = \(\frac{V}{R}\) =\(\frac{6}{6}\) = 1A

Question 12.
If each diode has a forward bias resistance of 25 ohms in the given circuit, then find the value of \(\frac{I_1}{I_2}\).

Answer:
In the figure diode D₂ is in reverse bias.
So current does not flow. Thus I₃ = 0

Objective Questions

Question 1.
When a triode is used as an amplifier the phase difference between the input signal voltage and the output is
1) 0
2) π
3) π/2
4) π/4
Answer:
2) π

Question 2.
In semiconductors at a room temperature
1) the valence band is partially empty and the conduction band is partially filled
2) the valence band is completely filled and the conduction band is partially filled
3) the valence band is completely filled
4) the conduction band is completely empty.
Answer:
1) the valence band is partially empty and the conduction band is partially filled

Question 3.
At absolute zero, Si acts as
1) non metal
2) metal
3) insulator
4) none of these.
Answer:
3) insulator

Question 4.
If a small amount of antimony is added to germanium crystal
1) it becomes a p-type semiconductor
2) the antimony becomes an acceptor atom
3) there will be more free electrons than holes in the semiconductor
4) its resistance is increased,
Answer:
3) there will be more free electrons than holes in the semiconductor

Question 5.
In a p type semiconductor, the majority carriers of current are
1) protons
2) electrons
3) holes
4) neutrons
Answer:
3) holes

Question 6.
Which of the following, when added us an impurity into the silicon produces n type semiconductor?
1) B
2) Al
3) P
4) Mg
Answer:
3) P

Question 7.
To obtain a p-type germanium semiconductor, it must be doped with
1) indium
2) phosphorus
3) arsenic
4) antimony.
Answer:
1) indium

Question 8.
When arsenic is added as an’ impurity to silicon, the’ resulting material is
1) n-type conductor
2) n-type semiconductor
3) p-type semiconductor
4) none of these.
Answer:
2) n-type semiconductor

Question 9.
When n type semiconductor is heated
1) number of electrons increases while that of holes decreases
2) number of holes increases while that of electrons decreases
3) number of electrons and holes remain same
4) number of electrons and holes increases equally.
Answer:
4) number of electrons and holes increases equally.

Question 10.
The increase in the width of the depiction region in a p-n junction diode is due to
1) forward bias only
2) reverse bias only
3) both forward bias and reverse bias
4) increase in forward current
Answer:
2) reverse bias only

Question 11.
In a p-n junction
1) high potential at n side and low potential at p side
2) high potential at p side and low potential at nside
3) p and n both are at same potential
4) undetermined.
Answer:
1) high potential at n side and low potential at p side

Question 12.
Depletion layer consists of
1) mobile ions
2) protons
3) electrons
4) immobile ions
Answer:
4) immobile ions

Question 13.
The depletion layer in the p-n junction region is caused by
1) drift of holes
2) diffusion of charge carriers
3) migration of impurity ions
4) drift of electrons.
Answer:
2) diffusion of charge carriers

Question 14.
Consider the junction diode as ideal. The value of current flowing through AB is

1) 10^-1 A
2) 10^-3 A
3) 0 A
4) 10^-2 A
Answer:
4) 10^-2 A

Question 15.
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is

1) 0.75 A
2) zero
3) 0.25 A
4) 0.5 A
Answer:
4) 0.5 A

Question 16.
Barrier potential of a p-n junction diode does not depend on
1) diode design
2) temperature
3) forward bias
4) doping density
Answer:
1) diode design

Question 17.
In forward bias, the width of potential barrier in a p-n junction diode
1) remains constant
2) decreases
3) increases
4) first (1) then (2)
Answer:
2) decreases

Question 18.
In a junction diode, the holes are due to
1) extra electrons
2) neutrons
3) protons
4) missing of electrons
Answer:
4) missing of electrons

Question 19.
The cause of the potential barrier in a p-n junction diode is
1) depletion of negative charges near the junction
2) concentration of positive charges near the junction
3) depletion of positive charges near the junction
4) concentration of positive and negative ch near the junction.
Answer:
4) concentration of positive and negative ch near the junction.

Question 20.
The peak voltage in the output of a half wave diode rectifier fed with a sinusoidal signal without filter is 10 V. The d.c. component of the output voltage is
1) 10/\(\sqrt{2}\)
2) 10/πV
3) 10 V
4) 20/π V
Answer:
2) 10/πV

Question 21.
If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
1) 25 Hz
2) 50Hz
3) 70.7 Hz
4) 100Hz
Answer:
4) 100Hz

Question 22.
A p-n junction diode can be used as
1) condenser
2) regulator
3) amplifier
4) rectifier
Answer:
4) rectifier

Question 23.
A Zener diode, having breakdown voltage equal to 15V, is used in a voltage regulator circuit shown in figure. The current through the diode is

1) 5 mA
2) 10 mA
3) 15 mA
4) 20 mA
Answer:
1) 5 mA

Question 24.
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
1) 4000 mn
2) 6000nm
3) 4000 A°
4) 6000 A°
Answer:
3) 4000 A°

Question 25.
Zener diode is used for
1) amplification
2) rectification
3) stabilisation
4) producing oscillations in an oscillator
Answer:
3) stabilisation

Question 26.
In the combination of the following gates the output Y can be written in terms of inputs A and B as

1) \(\overline{\mathrm{A} . \mathrm{B}}\)
2) \(\text { A. } \overline{\mathrm{B}}+\overline{\mathrm{A}} . \mathrm{B}\)
3) \(\overline{\mathrm{A} . \mathrm{B}}\) + A.B
4) \(\overline{A+B}\)
Answer:
2) \(\text { A. } \overline{\mathrm{B}}+\overline{\mathrm{A}} . \mathrm{B}\)

Question 27.
The given electrical network is equivalent to

1) OR gate
2) NOR gate
3) NOT gate
4) AND gate
Answer:
2) NOR gate

Question 28.
For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the terminal y?

Answer:

Question 29.
What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then I?

1) 0, 1
2) 0, 0
3) 1, 0
4) 1, 1
Answer:
3) 1, 0

Question 30.
The figure shows a logic circuit with two inputs A and B and the output C. The voltage wave forms across A, B and C are as given. The logic circuit gate is

1) OR gate
2) NOR gate
3) AND gate
4) NAND gate
Answer:
1) OR gate

Question 31.
Symbolic representation of four logic gates are shown as

Pick out which ones are for AND, NAND and NOT gate, respectively
1) (ii), (iii) and (iv)
2) (iii), (ii) and (i)
3) (iii), (ii) and (iv)
4) (ii), (iv) and (iii)
Answer:
4) (ii), (iv) and (iii)

Question 32.
The device that can act s a complele electron k circuit is
1) junction diode
2) integrated circuit
3) junction transistor
4) zener diode.
Answer:
2) integrated circuit

Question 33.
The following truth-table belongs to which one of the fo1loin tour gates?

1) XOR
2) NOR
3) OR
4) NAND
Answer:
2) NOR

Question 34.
This symbol represents

1) AND gate
2) NOR gate
3) NAND gate AB
4) OR gate
Answer:
3) NAND gate AB

Question 35.
The following truth table corresponds to the logical gate

1) NAND
2) OR
3) AND
4) XOR
Answer:
2) OR