WBCHSE Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions

WBCHSE Class 12 Physics Electromagnetism MCQs

Electromagnetism Multiple Choice Question And Answers

Question 1. Which of the properties of an isolated north pole, placed at a pointin’ field, is characterized by the direction of the tangent on the magnetic line of force passing through that point?

1. Position
2. Displacement
3. Velocity
4. Acceleration

Answer: 4. Acceleration

Question 2. Magnetic flux is defined as

1. The number of magnetic lines of force passing through a surface
2. The number of magnetic lines of force passing normally through a surface
3. The number of magnetic lines of force passing normally through the unit area of a surface
4. The number of magnetic Hnef-ice pa unit area of a surface

Answer: 2. The number of magnetic lines of force passing normally through a surface

Question 3. Current I is flowing through a vertical long wire in the upward direction. The magnetic field at a point, on the east of the wire is

1. Upwards
2. Towards north
3. Towards south
4. Towards west

Answer: 2. Towards north

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. A current of 1 A is flowing through a circular coil of radius 10 cm having N turns. If the magnetic field produced at the center of the coil is 4π x 10^-6 T, what is the value of N?

1. 20
2. 10
3. 2
4. 1

Answer: 3. 2

WBCHSE class 12 physics MCQs

Question 5. Which one le of the following relations expresses Biot-Savart law?

1. \(d \vec{B}=\frac{\mu_0 I}{4 \pi}: \frac{d l \times \vec{r}}{r^2}\)
2. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \vec{r}}{r^3}\)
3. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \vec{r}}{r}\)
4. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \hat{r}}{r}\)

Answer: 2. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \vec{r}}{r^3}\)

WBBSE Class 12 Electromagnetism MCQs

Question 6. If we double the radius of a current-carrying coil keeping the current unchanged, the magnetic field at its centre will

1. Remain Unchanged
2. Become Double
3. Be Halved
4. Become four times

Answer: 3. Be Halved

Question 7. A circular coil of radius r carries a current. It produces magnetic fields B1 at the center of the coil, and B₁ and B₂ at an axial point at a distance r from the center. The ratio of B₁ and B₂ is

1. √2: 1
2. 2: 1
3. 2√2: 1
4. 4: 1

Answer: 3. 2√2: 1

Question 8. A conductor is carrying a current I. The magnitude of the magnetic field at the origin is

1. \(-\frac{\mu_0 I}{4 r}\left(\frac{1}{\pi} \hat{i}+\frac{1}{2} \hat{k}\right)\)
2. \(\frac{\mu_0 I}{4 \pi}\left(\frac{1}{\pi} \hat{i}-\frac{1}{2} \hat{j}\right)\)
3. \(\frac{\mu_0 I}{4 r}\left(\frac{1}{2} \hat{i}-\frac{I}{R} \hat{j}\right) \frac{\mu_0}{4 r}\)
4. \(\frac{\mu_0 I}{4 r}\left(\frac{2}{\pi} \hat{i}+\hat{j}\right)\)

Answer: 1. \(-\frac{\mu_0 I}{4 r}\left(\frac{1}{\pi} \hat{i}+\frac{1}{2} \hat{k}\right)\)

Question 9. The ratio of magnetic fields at the center of a current-carrying coil of radius r and at a distance of 3 r on its axis is

1. V10
2. 2V10
3. 10V10
4. 20V10

Answer: 3. 10V10

WBCHSE class 12 physics MCQs

Question 10. Two wires PQ and QR carry equal currents I. One end of each wire extends to infinity and ∠PQR = θ. The magnitude of the magnetic field at O on the bisector of angle ∠PQR at a distance r from point Q is

1. \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \sin \frac{\theta}{2}\)
2. \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \cot \left(\frac{\theta}{2}\right)\)
3. \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \tan \frac{\theta}{2}\)
4. \(\frac{\mu_0}{2 \pi} \cdot \frac{I}{r}\left(\frac{1+\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right)\)

Answer: 4. \(\frac{\mu_0}{2 \pi} \cdot \frac{I}{r}\left(\frac{1+\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right)\)

Question 11. A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it at a small distance and point Q lies to the west at the same distance. The strength of the magnetic field at P is

1. Greater than that at Q
2. Same as that at Q
3. Less than that at Q
4. Greater or less depends on the strength of the current

Answer: 2. Same as that at Q

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE 

Question 12. The magnitude of the magnetic field at the point P in the arrangement.

1. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\left(1-\frac{1}{\sqrt{2}}\right)\)
2. \(\frac{2 \mu_0 i}{\sqrt{2} \pi d}\)
3. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\)
4. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\left(1+\frac{1}{\sqrt{2}}\right)\)

Answer: 1. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\left(1-\frac{1}{\sqrt{2}}\right)\)

Question 13. The magnetic field at point O due to the current I flowing through the wire, is

1. \(\frac{2 \mu_0 I}{r}\)
2. \(\frac{\mu_0 I}{r}\)
3. \(\frac{\mu_0 I}{2 r}\)
4. \(\frac{\mu_0 I}{4 r}\)

Answer: 4. \(\frac{\mu_0 I}{4 r}\)

Question 14. The magnetic field at the point of intersection of diagonals of a square wire loop of side L carrying a current I is

1. \(\frac{\mu_0 I}{\pi L}\)
2. \(\frac{2 \mu_0 I}{\pi L}\)
3. \(\frac{\sqrt{2} \mu_0 I}{\pi L}\)
4. \(\frac{2 \sqrt{2} \mu_0 I}{\pi}\)

Answer: 4. \(\frac{2 \sqrt{2} \mu_0 I}{\pi}\)

Question 15. A circular coil of radius R carries an electric current i. The magnetic field at a point on the axis at a distance x from, the center of the coil (x >> k) varies as

1. \(\frac{1}{x}\)
2. \(\frac{1}{x^{3 / 2}}\)
3. \(\frac{1}{x^2}\)
4. \(\frac{1}{x^3}\)

Answer: 4. \(\frac{1}{x^3}\)

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE 

Question 16. The magnetic field B within a solenoid of length I with N turns and carrying a current i is given by

1. \(\frac{\mu_0 N i}{e L}\)
2. \(\mu_0 N i\)
3. \(\frac{\mu_0^* N i}{L}\)
4. \(\frac{4 \pi \mu_0 N i}{L}\)

Answer: 3. \(\frac{4 \pi \mu_0 N i}{L}\)

Question 17. In a current-carrying, long solenoid, the field produced inside the solenoid does not depend upon

1. The radius of the solenoid
2. Number of turns per unit length
3. Current flowing through it
4. The medium in which the solenoid is placed

Answer: 1. Radius of the solenoid

Common MCQs on Electromagnetic Theory

Question 18. A charged particle enters a magnetic field \(\vec{B}\) perpendicularly with velocity v and keeps rotating along a circular path of radius r. What will happen if the magnitude of \(\vec{B}\) is increased?

1. v will increase
2. v will decrease
3. r will increase
4. r will decrease

Answer: 4. r will decrease

Question 19. The radius of the circular path described by a charged particle in a magnetic field is directly proportional to the

1. Momentum of the particle
2. The kinetic energy of the particle
3. Amount of charge of the particle
4. Strength of the magnetic field

Answer: 1. Momentum of the particle

Question 20. The magnitude of an electric field along the x-axis is 1 V .m^-1 and in the same region the magnitude of a magnetic field along the y-axis is 10^-6 T . What should be the velocity of an electron in that region so that it will continue to move with uniform velocity along z-axis without suffering any deviation?

1. 10⁶ m .s^-1
2. 10^-6 m.s^-1
3. 2 x 10⁶ m.s^-1
4. 2 x 10^-6 m.s^-1

Answer: 1. 10⁶ m.s^-1

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE

Question 21. A beam of protons projected along a positive X-axis experiences a force, due to a magnetic field, along the negative Y-axis. Then the magnetic field must be

1. Along the z-axis
2. Along the negative z-axis
3. On the xy-plane
4. On the xz-plane

Answer: 4. On the xz -plane

Question 22. A moving electron and a moving proton enter a uniform magnetic field in a direction perpendicular to that of the field. If the radii of their circular orbits are equal, they have the same

1. Velocity
2. Momentum
3. Kinetic energy
4. Charge to mass ratio

Answer: 2. Momentum

Question 23. A particle and a proton having the same momentum enter a region of a uniform magnetic field and move in circular paths. The ratio of the radii of curvature of their circular paths ra/rp in the field is

1. 1
2. 1/4
3. 1/2
4. 4

Answer: 3. 1/2

Question 24. A uniform electric field and a uniform magnetic field are acting along die same direction in a certain region. If an electron is projected in the region with a velocity along the direction of fields, then

1. The electron will turn toward the right
2. The speed of the electron will decrease
3. The speed of the electron will increase
4. The electron will turn toward the left

Answer: 2. The speed of the electron will decrease

Question 25. A particle of charge q moves with a velocity \(\vec{v}=a \hat{i}+b \hat{j}\) in a magnetic field \(\vec{B}=c \hat{i}+d \hat{j}\). The force acting on the particle has magnitude F. Then

1. F = 0 if ad = bc
2. F = 0 if ad = -bc
3. F = 0 if ac = -bd
4. \(F \propto\left(a^2+b^2\right)^{1 / 2} \times\left(c^2+d^2\right)^{1 / 2}\)

Answer: 1. F = 0 if ad = bc

Question 26. A proton, a deuteron, and an a -particle are accelerated by the same potential and then enter a uniform magnetic field perpendicularly. The ratio of radii of their circular paths will be

1. 1:√2:√2
2. 2: 2: 1
3. 1:2:1
4. 1: 1: 1

Answer: 1. 1:√2:√2

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE 

Question 27. Through a straight conducting wire, the current is flowing along a positive z-direction. What should be the direction of the applied magnetic field so that the wire will experience the maximum force?

1. Along the positive or negative z-axis
2. Along any direction, the xz-plane
3. Along any direction on xy-plane
4. Along any direction on Yz-plane

Answer: 3. Along any direction on xy-plane

Electric energy and power class 12 MCQs

Question 28. A conducting circular loop of radius r carries a constant current I. It is placed in a uniform magnetic field \(\vec{B}\) such that \(\vec{B}\) is perpendicular to the plane of the loop. The magnetic force acting on the loop is

1. BIr
2. 2πrIB
3. Zero
4. πrIB

Answer: 3. Zero

Practice Questions on Magnetic Fields and Forces

Question 29. A magnetic field is applied along the positive z-axis. How should a plane conducting loop be placed in this field so that the loop will not experience any torque?

1. On xy-plane
2. On xz-piane
3. On yz -plane
4. Along z-axis

Answer: 1. On xy-plane

Question 30. The path of a charged particle whose motion is perpendicular to a uniform magnetic field is

1. A straight line
2. An ellipse
3. A circle
4. A helix

Answer: 3. A circle

Question 31. Two concentric- coils each of radius 2π. cm are placed at right angles to each other. 3 A and 4A are die currents flowing in the coils. The magnetic induction in weber/m² at the centre of the coils will be (\(\mu_0=4 \pi \times 10^{-7} \mathrm{H} \cdot \mathrm{m}^{-1}\))

1. 5 x 10^-5
2. 7 x 10^-5
3. 12 x 10^-5
4. 10^-5

Answer: 1. 5 x 10^-5

Question 32. A square loop carrying a steady current I is placed on a horizontal plane near a long straight conductor carrying a steady current I1 at a distance. The loop will experience

1. A net attractive force toward the conductor
2. A net repulsive force away from the conductor
3. A net torque acting upwards perpendicular to the horizontal plane
4. A net torque acting downwards normal to the

Answer: 1. A net attractive force toward the conductor

WBCHSE Physics Electromagnetism Chapter MCQs 

Electric energy and power class 12 MCQs

Question 33. A square loop of side a is placed at a distance from a long wire carrying a current. If the loop carries a current I₂, then the nature of force and its magnitude is

1. \(\text { attractive, } \frac{\mu_0 I_1 I_2}{2 \pi a}\)
2. \(\text { attractive, } \frac{\mu_0 I_1 I_2}{4 \pi}\)
3. \(\text { repulsive, } \frac{\mu_0 I_1 I_2}{4 \pi}\)
4. \(\text { repulsive, } \frac{\mu_0 I_1 I_2}{4 \pi a}\)

Answer: 2. \(\text { attractive, } \frac{\mu_0 I_1 I_2}{4 \pi}\)

Question 34. For 1 A current, a galvanometer shows its full-scale deflection. If a resistance of 800Ω is connected in series, it is converted into a voltmeter of range 0-1000 V. What is the resistance of the galvanometer?

1. 50Ω
2. 100Ω
3. 200Ω
4. 800Ω

Answer: 3. 200Ω

Question 35. In an ammeter, 0.5% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, the resistance of the ammeter will be

1. \(\frac{G}{200}\)
2. \(\frac{G}{199}\)
3. 200 G
4. 199 G

Answer: 1. \(\frac{G}{200}\)

Question 36. Two charged particles traverse identical helical paths completely opposite sense in a uniform magnetic field, \(\vec{B}=B_0 \hat{k}\)

1. They have equal z-components of momenta
2. They must have equal charge
3. They necessarily represent particle-antiparticle pair
4. The charge to mass ratio satisfy \(\left(\frac{q}{m}\right)_1+\left(\frac{q}{m}\right)_2=0\)

Answer: 4. The charge to mass ratio satisfy \(\left(\frac{q}{m}\right)_1+\left(\frac{q}{m}\right)_2=0\)

The radius of the helical path, \(r=\frac{\nu \sin \theta}{\left(\frac{q}{m}\right) B}\)

and the displacement of the charged particle along the magnetic field is one rotation

⇒ \(\frac{2 \pi \nu \cos \theta}{\left(\frac{q}{m}\right) B}\)

Two charged particles will traverse identical helical paths in the opposite sense is \(\frac{q}{m}\) for both the particles is same and opposite.

WBCHSE Physics Electromagnetism Chapter MCQs 

Question 37. An electron is projected with uniform velocity along the axis of a current carrying a long solenoid. Which one of the following is true?

1. The electron will be accelerated along the axis
2. The electron path will be circular about the axis
3. The electron will experience a force at 45° to the axis and execute a helical path
4. The electron will continue to move with uniform velocity along the axis of the solenoid

Answer: 4. The electron will continue to move with uniform velocity along the axis of the solenoid

F = evBsinθ

As θ= 0, F = 0.

Important Definitions in Electromagnetism

Question 38. Consider a wire carrying a steady current I placed in a uniform magnetic field \(\vec{B}\). Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that

1. The motion of charges inside the wire moves to the surface as a result of \(\vec{B}\).
2. Some charges inside the wire move to the surface as a result of \(\vec{B}\)
3. If the wire moves under the influence of \(\vec{B}\), no work is done by the force
4. If the wire moves under the influence of \(\vec{B}\), no work is done by the magnetic force on the ions, assumed to be fixed within the wire.

Answer:

2. Some charges inside the wire move to the surface as a result of \(\vec{B}\)

4. If the wire moves under the influence of \(\vec{B}\), no work is done by the magnetic force on the ions, assumed to be fixed within the wire.

⇒ \(\vec{F}=q(\vec{v} \times \vec{B}),\)

⇒ \(\vec{F}=I(\vec{l} \times \vec{B})\)

The direction of force due to the magnetic field is perpendicular to the displacement of the charge.

Question 39. A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity \(\vec{v}\) and a position enters via the opposite face with velocity \(\vec{v}\). At this instant,

1. The electric forces on both particles cause identical accelerations
2. The magnetic forces on both particles cause equal acceleration
3. Both particles gain or lose energy at the same rate
4. The motion of the center of mass is determined by \(\vec{B}\) alone

Answer:

2. The magnetic forces on both the particles cause equal acceleration

3. Both particles gain or lose energy at the same rate

4. The motion of the center of mass is determined by \(\vec{B}\) alone

For electric field, \(\vec{F}=q \vec{E}\)

So, \(\vec{F}_e=-e \vec{E} \text { and } \vec{F}_p=+e \vec{E}\)

Resultant electric force \(\vec{F}_e+\vec{F}_p=\overrightarrow{0}\)

For magnetic field, \(\vec{F}_e=-e \vec{v} \times \vec{B}\)

⇒ \(\vec{F}_p=+e(-\vec{v} \times \vec{B})=-e \vec{v} \times \vec{B}=\vec{F}_e\)

Class 12 physics electric energy questions

Question 40. A charged particle would continue to move with a constant velocity in a region where

1. \(\vec{E}=\overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)
2. \(\vec{E} \neq \overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)
3. \(\vec{E} \neq \overrightarrow{0}, \vec{B}=\overrightarrow{0}\)
4. \(\vec{E}=\overrightarrow{0}, \vec{B}=\overrightarrow{0}\)

Answer:

1. \(\vec{E}=\overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)

2. \(\vec{E} \neq \overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)

4. \(\vec{E}=\overrightarrow{0}, \vec{B}=\overrightarrow{0}\)

No force acts; if \(\vec{v} \text { and } \vec{B}\) are parallel.

When \(\vec{v}, \vec{E} \text { and } \vec{B}\) are mutually perpendicular, \(\vec{v}\) does not change if \(\frac{E}{B}=v\).

For force \(q \vec{E}(\neq 0), \vec{v}\) must change.

No force acts; so \(\vec{v}\) does not change.

Question 41. Two identical current-carrying coaxial loops, carry current I in an opposite sense. A simple campervan loop passes through both of them once. Calling the loop as C,

1. \(\oint_C \vec{B} \cdot d \vec{l}=\mp 2 \mu_0 I\)
2. The value of \(\int_C \vec{B} \cdot d \vec{l} \text { is }\) independent of the shape of C
3. There may be a point on C where \(\vec{B} \text { and } d \vec{l}\) is perpendicular
4. \(\vec{B}\) vanishes everywhere on C

Answer:

2. The value of \(\int_C \vec{B} \cdot d \vec{l} \text { is }\) independent of the shape of C

3. There may be a point on C where \(\vec{B} \text { and } d \vec{l}\) is perpendicular

⇒ \(\oint_C \vec{B} \cdot d \vec{l}=\mu_0(I-I)=0\) \(\left[∵ \oint_C \vec{B} \cdot d \vec{l}=\mu_0 I\right]\)

At the two extremities of C, \(\vec{B}\) and \(\vec{dl}\) are perpendicular to each other.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 42. A magnetic needle facing north-south can rotate freely in a horizontal plane. A conducting wire is placed parallel to it in a north-south direction.

1. The direction of the current is from south to north and the conductor is above the magnet-north pole of the magnet will be deflected towards west.
2. The direction of the current is from north to south, the conductor is above the magnet-south pole of the magnet will be deflected towards the west.
3. The direction of the current is from south to north, the conductor is below the magnet-the south pole of the magnet will be deflected towards west.
4. The direction of the current is from north to south, the conductor is below the magnet-the north pole of the magnet will be deflected towards west.

Answer:

1. The direction of the current is from south to north and the conductor is above the magnet-north pole of the magnet will be deflected towards west.

2. The direction of the current is from north to south, the conductor is above the magnet-south pole of the magnet will be deflected towards the west.

3. The direction of the current is from south to north, the conductor is below the magnet south pole of the magnet will be deflected towards the west.

4. The direction of the current is from north to south, the conductor is below the magnet north pole of the magnet will be deflected towards the west.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 43. The magnetic field developed at a point near a straight-carrying conductor depends on

1. Material of the conductor
2. Distance of the point from the conductor
3. Direction and magnitude of current
4. Medium between the point and conductor

Answer:

2. Distance of the point from the conductor

3. Direction and magnitude of current

4. Medium between the point and conductor

Examples of Electromagnetic Induction Questions

Question 44. A current I passes through the conductor ABCD, The magnetic field at 0 In the same plane is

1. Zero For the currents in segments AD and CD
2. Zero for current in segment BC
3. \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) for the currents in segments AB and CD
4. \(\frac{\mu_0 I \theta}{4 \pi r}\) for current in segment BC

Answer:

1. Zero For the currents in segments AD and CD

4. \(\frac{\mu_0 I \theta}{4 \pi r}\) for current in segment BC

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 45. Two identical charged particles enter a uniform magnetic field with the same speed but at angles 30° and 60° with the field. Let a, b, and c be the ratios of their time periods, radii, and pitches of the helical paths. Then

1. abc = 1
2. abc > 1
3. abc < 1
4. a = bc

Answer:

1. abc = 1

4. a = bc

Class 12 physics electric energy questions

Question 46. In case of a current carrying solenoid

1. Internal lines of force are parallel
2. The lines of force become congested upon increasing the current
3. No north-south pole is produced in the absence of a core of magnetic material
4. The magnetic field increases on increasing the number of turns

Answer:

1. Internal lines of force are parallel

2. The lines of force become congested upon increasing the current

4. The magnetic field increases on increasing the number of turns

Question 47. A particle with charge q is moving with a velocity \(\vec{v}\) in a magnetic field \(\vec{B}\). If the force acting on the particle is \(\vec{F}\) then,

1. \(\vec{F}=0 \text {, when } \vec{v} \text { and } \vec{B}\) are parallel
2. Magnitude of \(\vec{F}\) is maximum when \(\vec{v} \text { and } \vec{B}\) are perpendicular to each other
3. \(|\vec{F}|=\frac{1}{2} q v B\) when the angle between \(\vec{v} \text { and } \vec{B}\) is 45°
4. The direction of \(\vec{F}\) remains the same when the directions of both \(\vec{v} \text { and } \vec{B}\) are reversed simultaneously

Answer:

1. \(\vec{F}=0 \text {, when } \vec{v} \text { and } \vec{B}\) are parallel

2. Magnitude of \(\vec{F}\) is maximum when \(\vec{v} \text { and } \vec{B}\) are perpendicular to each other

4. Direction of \(\vec{F}\) remains the same when the directions of both \(\vec{v} \text { and } \vec{B}\) are reversed simultaneously

Question 48. The rotation of Barlow’s wheel is due to the action of the magnetic field on the current. In the case of this wheel.

1. The rotational speed does not increase if both the current and the magnetic field are increased simultaneously
2. The rotational speed increases when the current is increased
3. The rotational speed increases when the magnetic field is increased
4. The rotational speed becomes higher as the wheel is made lighter

Answer:

2. The rotational speed increases when the current is increased

3. The rotational speed increases when the magnetic field is increased

4 The rotational speed becomes higher as the wheel is made lighter

Question 49. Two loan straight conducting wires arc kept parallel to one other at a distance r apart. When a current I passes through both the wires In the same direction, a force of attraction F acts between the wires. Which of the following statements is/are true?

1. When r – 0.5 m and I = 1 A, F = 10^-7 N
2. When r = 2.0 m and I = 1 A , F = 10^-7 N
3. When r = 8.0 m and I = 2 A, F = 10^-7 N
4. When r = 1.0 m and I = 1 A, F = 2 x 10^-7 N

Answer:

1. When r – 0.5 m and I = 1 A, F = 10^-7 N

2. When r = 2.0 m and I = 1 A , F = 10^-7 N

4. When r = 1.0 m and I = 1 A, F = 2 x 10^-7 N

Electric power multiple-choice questions

Question 50. Equal and opposite currents flow through two long straight and parallel wires. A’, B, and C are three points on the same plane.

1. The magnetic field developed at B is zero
2. The magnetic field developed at B is maximum and acts downward with respect to the plane
3. The magnetic field developed at A acts upwards with respect to the plane
4. The magnetic field developed at C acts downwards with respect to the plane

Answer:

2. The magnetic field developed at B is maximum and acts downward with respect to the plane

3. Magnetic field developed at A acts upwards with respect to the plane

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 51. Usually, the resistance G of a galvanometer is of moderate value. But it has a full-scale deflection for a very small current I_G. In the state of full-scale deflection the terminal potential difference of the galvanometer V_G = G.I_G. Usually this V_G is also very small. If the current in an electrical circuit I > IG and the galvanometer is directly connected in series with the circuit, it cannot measure the current through the circuit; moreover, there is the possibility of the galvanometer being damaged as a result of high current. Now if a resistor of small resistance S is connected in parallel with the galvanometer as a shunt, a major portion of the current flows through the shunt. If the galvanometer shows full deflection for a current I (I > I_G) in the main circuit, then it can be said that the galvanometer is capable of measuring a current up to I i.e., it will become an ammeter of range up to a current I. On the other hand, if the potential difference between two points of the main circuit V > V_G and the galvanometer is connected in parallel with the circuit it cannot measure the potential difference V; moreover since V > V_G, the galvanometer may be damaged. Now if a resistor of high resistance R is connected in series with the galvanometer, the potential difference between the two terminals A and B of the combination is much larger than VG. If this combination shows full-scale deflection for a potential difference V (V > V_G) when connected in parallel to two points of the main circuit, then it can be said that the galvanometer is capable of measuring a potential difference up to V, i.e., it has now become a voltmeter of range up to V.

1. The resistance of the shunt to be connected with the galvanometer to convert it to an ammeter capable of measuring a current of nlG (n > 1) is

1. nG
2. \(\frac{G}{n+1}\)
3. \(\frac{G}{n}\)
4. \(\frac{G}{n-1}\)

Answer: 4. \(\frac{G}{n-1}\)

Electric power multiple-choice questions

2. The resistance to be connected in series with this galvanometer to convert it to a voltmeter capable of measuring a potential difference of nVQ (n > l) is

1. nG
2. (n+1)G
3. (n-1)G
4. \(\frac{n}{n-1} G\)

Answer: 3. (n-1)G

3. The range of the ammeter was designed by connecting a 10a shunt with a 100Ω – 100μ A galvanometer is

1. 0-1 mA
2. 0-1.1 mA
3. 0-10 mA
4. 0-11 mA

Answer: 2. 0-1.1 mA

4. A galvanometer of 500Ω is converted to an ammeter of range 0- 10 A when a 5Ω shunt is connected to it. The maximum current that can flow through the galvanometer is

1. 98 mA
2. 99 mA
3. 100 mA
4. 101 mA

Answer: 2. 99 mA

5. A 10Ω galvanometer shows full-scale deflection for a current of 10 mA. What resistance should be connected with it in series so as to convert it to a voltmeter of range 0- 5 V.

1. 490Ω
2. 499Ω
3. 500Ω
4. 510Ω

Answer: 490Ω

Electromagnetism MCQs for Class 12 Physics WBCHSE 

6. The resistance of a voltmeter is 300Ω. It shows full-scale deflection for a terminal potential difference of 150 V. What resistance should be connected in parallel with the voltmeter to convert it to an ammeter of range θ = 8 A?

1. 19Ω
2. 20Ω
3. 21Ω
4. 190Ω

Answer: 2. 20Ω

7. When a resistance R is connected in parallel to a galvanometer, the range of current measurable by it becomes n times. On the other hand, if the same resistance is connected in series, the range of potential difference measurable by it becomes n’ times. The value of n’ is

1. n-1
2. n+1
3. \(\frac{n+1}{n}\)
4. \(\frac{n}{n-1}\)

Answer: 4. \(\frac{n}{n-1}\)

Question 52. The direction of the magnetic field developed at a point near a long straight current-carrying wire can be determined by Mazwell’s cork screw rule. This rule stated that if a right-handed corkscrew is driven along the direction of current in a conductor, the direction of rotation of the thumb at any point near the conductor gives the direction of the magnetic field at that point. If the current through such a conductor is I, then the value of magnetic field developed at a distance r from the conductor in any directionis given by \(B=\frac{\mu_0 I}{2 \pi r}\), where \(\mu_0=4 \pi \times 10^{-7} \mathrm{H} \cdot \mathrm{m}^{-1}\) and unit of B is tesla (T). Let us now consider two long straight conducting wires 1 and 2 kept rigidly at a distance of 4 cm and parallel to each other. A current of 10 A flows through each wire in the same direction. In P and Q are two points on the plane of the wires and both are 1 cm away from the first wire.

1. The direction of the resultant magnetic field at P is

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward left
4. Toward right

Answer: 2. Upward with respect to the page

2. Value of resultant magnetic field at P (in T)

1. 1.33 x 10^-4
2. 1.6 x 10^-4
3. 2.4 x 10^-4
4. 2.67 x 10^-4

Answer: 3. 2.4 x 10^-4

WBCHSE physics electric energy MCQs 

3. Direction of the resultant magnetic field at Q

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward right
4. Toward left

Answer: 1. Downward with respect to the page

4. Value of resultant magnetic field at Q (in T)

1. 1.33 x 10^-4
2. 1.6 x 10^-4
3. 2.4 x 10^-4
4. 2.67 x 10^-4

Answer: 1. 1.33 x 10^-4

5. The direction of current is reversed in conductor 2, i.e., the current in the two wires are now in opposite directions, then the resultant magnetic field at P is

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward left
4. Toward right

Answer: 2. Upward with respect to the page

Electromagnetism MCQs for Class 12 Physics WBCHSE 

6. In condition (v) the value of the resultant magnetic field at P (in T)

1. 1.33 x 10^-4
2. 1.6 x 10^-4
3. 2.4 x 10^-4
4. 2.67 x 10^-4

Answer: 2. 1.6 x 10^-4

7. In condition (v) the direction of the resultant magnetic field at Q is

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward left
4. Toward right

Answer: 1. Downward with respect to the page

Question 53. A circular loop of radius a and two long parallel wires (marked I and II) are all in the plane of the paper. The distance of each wire from the center of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

1. When d≈a, but the wires are not touching the loop it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case

1. Current in wire I and wire II is in the direction PQ and RS, respectively, and h ≈ a
2. Current in wire I and wire II is in the direction PQ and SR, respectively, and h ≈ a
3. Current in wire I and wire II is in the direction PQ and SR respectively and h≈1.2a
4. Current in wire I and wire II is in the direction PQ and RS respectively and h ≈ 1.2a

Answer: 3. Current in wire I and wire II is in the direction PQ and SR respectively and h≈1.2a

Electromagnetism MCQs for Class 12 Physics WBCHSE 

2. Consider d>>a and the loop is rotated about its diameter parallel to the wires by 30° from the position. If the current in the wires is in the opposite directions, the torque on the loop at its new position will be (consider that the net field due to the wires is constant over the loop)

1. \(\frac{\mu_0 I^2 a^2}{d}\)
2. \(\frac{\mu_0 I^2 a^2}{2 d}\)
3. \(\frac{\sqrt{3} \mu_0 1^2 a^2}{d}\)
4. \(\frac{\sqrt{3} \mu_0 1^2 a^2}{2 d}\)

Answer: 2. \(\frac{\mu_0 I^2 a^2}{2 d}\)

Question 54. There is a constant homogeneous electric field of 100 V.m^-1 within the region x = 0 and x = 0.167 m along the positive x-direction. There is a constant homogeneous magnetic field within the region x = 0.167 m and x = 0.334 m along the z-direction. A proton at rest at the origin (x = 0, y = 0) is released along the positive x-direction.

1. Acceleration produced in the proton is (in m.s^-2)

1. 9.58 x 10⁹ \(\hat{i}\)
2. 5.98 x 10⁹ \(\hat{i}\)
3. 8.95 x 10⁹ \(\hat{i}\)
4. 3.33 x 10⁸ \(\hat{i}\)

Answer: 9.58 x 10⁹ \(\hat{i}\)

WBCHSE physics electric energy MCQs 

2. The path described by the proton in the magnetic field is

1. Straight
2. Circular
3. Parabolic
4. Elliptical

Answer: 2. Circular

3. The minimum value of the magnetic field (\(\vec{B}\)) necessary to confirm the proton within the region x = 0.167 m to x = 0.334 m is

1. 5 X 10^-1 T
2. 8 X 10^–4 T
3. 6 X 10^-2 T
4. 7 X 10^-3 T

Answer: 4. 7 X 10^-3 T

Question 55. A proton with a speed of 2 x 10⁷ m.s^-1 enters a magnetic field of flux density 1.5 Wb.m^-2, making an angle of 30° with the field. The force acting on the proton is

1. 2.4 x 10^-14 N
2. 0.24 x 10^-12 N
3. 0.024 x 10^-24 N
4. 24 x 10^-12 N

Answer: 1. 2.4 x 10^-14 N

We know, \(|F|=q v B \sin \theta\)

given, B = 1.5 Wb.m^-2, v = 2 x 10⁷ m.s^-1, 6 = 30°

q = 1.6 x 10^-19 C

Then, \(F=1.6 \times 10^{-19} \times 2 \times 10^7 \times 1.5 \times \frac{1}{2}=2.4 \times 10^{-12} \mathrm{~N}\)

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 56. A straight conductor of length l m carrying a current IA is bent in the form of a semicircle. The magnetic field (in tesla) at the center of the semicircle is

1. \(\frac{\pi^2 I}{l} \times 10^{-7}\)
2. \(\frac{\pi I}{l} \times 10^{-7}\)
3. \(\frac{\pi I}{l^2} \times 10^{-7}\)
4. \(\frac{\pi R^2}{l} \times 10^{-7}\)

Answer: 1. \(\frac{\pi^2 I}{l} \times 10^{-7}\)

Ifradius of the semicircle is r,

then, \(\pi r=l \quad \text { or, } r=\frac{l}{\pi}\)

Now, the magnetic field at the center of the semicircle,

⇒ \(B=\frac{\mu_0 I}{4 r}=\frac{\mu_0 I \pi}{4 l}=\frac{\mu_0}{4 \pi} \cdot \frac{\pi^2 I}{l}\)

⇒ \(\frac{\pi^2 I}{l} \times 10^{-7}\left[∵ \frac{\mu_0}{4 \pi}=10^{-7}\right]\)

The option 1 is correct

Question 57. A galvanometer having Internal resistance 10Ω requires 0.01 A for a full-scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at 120 V, we need to connect a resistance of

1. 11990Ω in series
2. 11990Ω in parallel
3. 12010Ω in series
4. 12010Ω in parallel

Answer: 1. 11990Ω in series

The internal resistance of the galvanometer (G) is 10Ω. A high-resistance R is to be added in series with a galvanometer. In this case, if the galvanometer is connected between the two points A and B then the galvanometer will become a voltmeter of voltage range from 0 to (V_A – V_B).

V_A – V_B = I_G(G+ R)

∴ \(R=\frac{V_A-V_B}{I_G}-G\)

⇒ \(\frac{120}{0.01}-10\)

=12000-10

= 11990Ω

The option 1 is correct.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 58. A long conducting wire carrying a current I is bent at 120°. The magnetic field B at a point on the right bisector of the bending angle at a distance d from the bend is (μ₀ is the permeability of free space)

1. \(\frac{3 \mu_0 I}{2 \pi d}\)
2. \(\frac{\mu_0 I}{2 \pi d}\)
3. \(\frac{\mu_0 I}{\sqrt{3} \pi d}\)
4. \(\frac{\sqrt{3} \mu_0 I}{2 \pi d}\)

Answer: 4. \(\frac{\sqrt{3} \mu_0 I}{2 \pi d}\)

⇒ \(P Q=d, P R=\frac{d \sqrt{3}}{2}=P S\)

The magnetic field at P due to the portion QT,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{\frac{d \sqrt{3}}{2}}\left(\sin 90^{\circ}+\sin 30^{\circ}\right)\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{d \sqrt{3}} \cdot \frac{3}{2}=\frac{\sqrt{3} \mu_0 I}{4 \pi d}\)

Magnetic field at P due to the portion \(Q U, B_2=\frac{\sqrt{3} \mu_0 I}{4 \pi d}\)

∴ Total magnetic field at \(P=B_1+B_2=\frac{\sqrt{3} \mu_0 I}{2 \pi d}\)

The option 4 is correct

WBCHSE physics electric energy MCQs 

Question 59. A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists a uniform magnetic field B, perpendicular to the plane of the motion, the proton will move in a circular path of radius

1. \(\frac{2 E m}{q B}\)
2. \(\frac{\sqrt{2 E m}}{q B}\)
3. \(\frac{\sqrt{E m}}{2 q B}\)
4. \(\sqrt{\frac{2 E q}{m B}}\)

Answer: 2. \(\frac{\sqrt{2 E m}}{q B}\)

The kinetic energy of the proton,

⇒ \(E=\frac{1}{2} m v^2=\frac{1}{2} m \frac{q^2 B^2 r^2}{m^2}\)

⇒ \(\frac{q^2 B^2 r^2}{2 m}\left[∵ \text { Velocity of the proton, } v=\frac{q B r}{m}\right]\)

∴ \(r^2=\frac{2 m E}{q^2 B^2} \quad \text { or, } r=\frac{\sqrt{2 E m}}{q B}\)

The option 2 is correct.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 60. A stream of electrons and protons are directed towards a narrow slit on a screen. The intervening region has a uniform electric field \(\vec{E}\) (vertically downwards) and a uniform magnetic field \(\vec{B}\).

1. Electrons and proton with speed \(\frac{|\vec{E}|}{|\vec{B}|}\) will pass through the slit
2. Protons with speed \(\frac{|\vec{E}|}{|\vec{B}|}\) will pass through the slit, electrons of the same speed will not
3. Neither electrons nor proton will go through the slit irrespective of their speed
4. Electrons will always be deflected upwards irrespective of their speed

Answer: 3 and 4 are correct

Let the directions of \(\vec{E}, \vec{v} \text { and } \vec{B}\) along the x-axis, y-axis, and z-axis respectively.

The electric force +eE and magnetic force +evB both act on the proton along the x-axis. So the proton will be deflected from the y-axis. Similarly, as the electron is negatively charged the two forces will deflect the electron along the negative x-axis. So neither the electron nor the proton will go through the slit.

The options 3 and 4 are correct.

Question 61. Two particles A and B, having equal charges, after being accelerated through the same potential difference enter a region of uniform magnetic field and the particles describe circular paths of radii R₁ and R₂ respectively. The ratio of the masses of A and B is

1. \(\sqrt{R_1 / R_2}\)
2. \(R_1 / R_2\)
3. \(\left(R_1 / R_2\right)^2\)
4. \(\left(R_2 / R_1\right)^2\)

Answer: 3. \(\left(R_1 / R_2\right)^2\)

For a circular path, magnetic force = centripetal force

i.e., \(q v B=\frac{m v^2}{R}\)

again, \(\frac{1}{2} m v^2=q V\)

hence, \(q v B=\frac{2 q V}{R}\)

or, \(R=\frac{2 V}{v B}=\frac{2 V}{B} \sqrt{\frac{m}{2 q V}}\left[∵ v=\sqrt{\frac{2 q V}{m}}\right]\)

For a circular path, magnetic force = centripetal force.

Therefore, \(\frac{R_1}{R_2}=\sqrt{\frac{m_1}{m_2}} \text { or, } \frac{m_1}{m_2}=\left(\frac{R_1}{R_2}\right)^2\)

The option 3 is correct.

Question 62. A rectangular coil carrying current is placed in a nonuniform magnetic field. On that coil the total

1. Force is non-zero
2. Force is zero
3. Torque is zero
4. Torque is non-zero

Answer: 1 and 4 are correct

If a current-carrying coil is placed in a non-uniform magnetic field, then, in general, both net force and net torque are non-zero. The options 1 and 4 are correct.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 63. The magnetic field due to a current in a straight wire segment of length L at a point on its perpendicular bisector at a distance r(r>>L)

1. \(\text { decreases as } \frac{1}{r}\)
2. \(\text { decreases as } \frac{1}{r^2}\)
3. \(\text { decreases as } \frac{1}{r^3}\)
4. \(\text { approaches a finite limit as } r \rightarrow \infty\)

Answer: 2. \(\text { decreases as } \frac{1}{r^2}\)

According to Biot-Savart law, \(d B \propto \frac{1}{r^2}\)

The option 2 is correct.

WBCHSE physics electric energy MCQs 

Question 64. The magnets of two suspended coil galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side a and that of the second one is circular of radius \(\frac{a}{\sqrt{\pi}}\). When the same current is passed through the coils, the ratio of the torque experienced by the first coil to that experienced by the second one is

1. \(1: \frac{1}{\sqrt{\pi}}\)
2. 1:1
3. π: 1
4. 1: π

Answer: 2. 1:1

In this case, applied torque (\(\tau\)) does not depend on the shape of the coil but on the area of cross-section (A) of the coil.

∴ \(\frac{\tau_1}{\tau_2}=\frac{A_1}{A_2}=\frac{a^2}{\pi\left(\frac{a}{\sqrt{\pi}}\right)^2}=1\)

or, \(\tau_1: \tau_2=1: 1\)

The option 2 is correct.

Question 65. A proton is moving with a uniform velocity of 10⁶ m.s^-1 along the F-axis, under the joint action of a magnetic field along the z-axis and an electric field of magnitude 2 x 10⁴ V.m^-1 along the negative X-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly (given: \(\frac{e}{m}\)ratio for proton ≈ 108 C.kg^-1 )

1. 0.5 m
2. 0.2 m
3. 0.1m
4. 0.05 m

Answer: 1. 0.5 m

Initially, the Lorentz force acting on the proton,

⇒ \(\vec{F}=\vec{F}_e+\vec{F}_m=\vec{E} q+q \vec{v} \times \vec{B}=-E q \hat{i}+q v B \hat{i}\)

Since the proton has no acceleration,

∴ \(\vec{F}=0 \quad ∴ E q=q v B\)

or, \(B=\frac{E}{v}=\frac{2 \times 10^4}{10^6}=2 \times 10^{-2} \mathrm{~T}\)

When the electric field is switched off the radius of the circular path,

⇒ \(R=\frac{m v}{q B}=\frac{10^6}{10^8 \times 2 \times 10^{-2}}=0.5 \mathrm{~m}\)

The option 1 is correct.

Question 66. Two long parallel wires separated by 0.1 m carry currents of 1A and 2A respectively in opposite directions. A third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of

1. 0.5 m from the 1st wire, towards the 2nd wire
2. 0.2 m from the 1st wire, towards the 2nd wire
3. 0.1 m from the 1st wire, away from the 2nd wire
4. 0.2 m from the 1st wire, away from the 2nd wire

Answer: 3. 0.1 m from the 1st wire, away from the 2nd wire

Let the 3rd wire be at a distance x from the first wire.

Since, \(B_1=B_2 \quad \text { or, } \frac{2 i_1}{x}=\frac{2 i_2}{(0.1+x)} \quad \text { or, } x=0.1 \mathrm{~m}\)

The option 3 is correct.

Real-Life Applications of Electromagnetic Principles

Question 67. A proton of mass m moving with a speed V << c, velocity of light in vacuum) completes a circular orbit in time T in a uniform magnetic field. If the speed of the proton is increased to √2v, what will be the time needed to complete the circular orbit?

1. √2
2. T
3. \(\frac{T}{\sqrt{2}}\)
4. \(\frac{T}{2}\)

Answer:

The magnetic force on a proton of mass m moving along a circular path in a uniform magnetic field,

⇒ \(\vec{F}_m=q(\vec{v} \times \vec{B})\)

∴ \(F_m=q v B \sin 90^{\circ}\) [B = magnetic field, q = charge of proton]

= qvB

Since the proton is moving along a circular path of radius r,

⇒ \(\frac{m v^2}{r}=q v B \quad .. v=\frac{q B r}{m}\)

Now the time taken by the proton to make one complete revolution is \(\frac{2 \pi r}{v}\)

Hence, \(T=\frac{2 \pi r}{v} \text { or, } T=\frac{2 \pi r \times m}{q B r} \text { or, } T=\frac{2 \pi m}{q B}\)

Since T does not depend on the velocity ofdie proton, if the velocity changes to V2v then T will remain unchanged.

The option 2 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 68. A uniform current is flowing along the length of an infinite, straight, thin, hollow cylinder of radius R. The magnetic field B produced at a perpendicular distance d from the axis of the cylinder is plotted in a graph. Which of the following figures looks like the plot?

Answer: 3.

Applying Ampere’s circuital law, a magnetic field at a point outside the cylinder and located at a perpendicular distance d from its axis,

⇒ \(B=\frac{\mu_0 I}{2 \pi d} \text { if, } d>R\)

⇒ \(\text { If } d=R \text { then, } B=\frac{\mu_0 I}{2 \pi R}\)

If d < R, B = 0 [since no current flows inside a hollow cylinder]

∴ \(B \propto \frac{1}{d}\)

The option 3 is correct.

WBCHSE physics electric energy MCQs 

Question 69. A circular loop of radius r of conducting wire connected with a voltage source of zero internal resistance produces a magnetic field B at its center. If instead, a circular loop of radius 2r, made of the same material, having the same cross-section is connected to the same voltage source, what will be the magnetic field at its center?

1. \(\frac{B}{2}\)
2. \(\frac{B}{4}\)
3. 2B
4. B

Answer: 2. \(\frac{B}{4}\)

The magnetic field at the center of a circular conducting loop,

⇒ \(B=\frac{\mu_0 I}{2 r}\) [I = current in the loop and r = radius of the loop]

If the resistance of the loop in the first case is R, then the
resistance in the second case,

R’ = 2R [since in the second case the length of the loop is doubled]

∴ In the second case, current in the loop, \(I^{\prime}=\frac{I}{2}\)

∴ \(B^{\prime}=\frac{\mu_0 I^{\prime}}{2 \times 2 r}=\frac{\mu_0 \times \frac{I}{2}}{2 r \times 2}=\frac{B}{4}\)

The option 2 is correct.

Question 70. A light-charged particle is revolving in a circle of radius r in the electrostatic attraction of a static heavy particle with an opposite charge. How does the magnetic field B at the center of the circle due to the moving charge depend on r?

1. \(B \propto \frac{1}{r}\)
2. \(B \propto \frac{1}{r^2}\)
3. \(B \propto \frac{1}{r^{3 / 2}}\)
4. \(B \propto \frac{1}{r^{5 / 2}}\)

Answer: 4. \(B \propto \frac{1}{r^{5 / 2}}\)

Let a light particle of mass m and charge -q2 revolve around a heavy particle of charge +q1 in a circular path of radius r with a velocity v

Now, \(\frac{m v^2}{r}=\frac{1}{4 \pi \epsilon_0} \frac{\left(+q_1\right) \times\left(-q_2\right)}{r^2}\)

or, \(v^2=\frac{1}{4 \pi \epsilon_0} \cdot \frac{-q_1 q_2}{m r}\)

∴ \(v \propto \frac{1}{\sqrt{r}}\)

The time period of the charge in a circular orbit,

⇒ \(T=\frac{2 \pi r}{v}\)

∴ Currently due to the revolving charge,

⇒ \(I=\frac{-q_2}{T}=\frac{q_2 \times v}{2 \pi r}\)

∴ \(I \propto \frac{v}{r} \quad \text { or, } I \propto \frac{1}{r^{3 / 2}}\left[∵ \nu \propto \frac{1}{\sqrt{r}}\right]\)

So, the magnetic field at the center of the circle due to the moving charge +q2,

⇒ \(B=\frac{\mu_0 I}{2 r} \quad \text { or, } B \propto \frac{I}{r}\)

or, \(B \propto \frac{1}{r^{3 / 2} \times r} \quad \text { or, } B \propto \frac{1}{r^{5 / 2}}\)

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 71. A conductor lies along the z-axis at -1.5 < z < 1.5 m and carries a fixed current of 10.0 A in the -ve z-direction. For a field \(B=3.0 \times 10^{-4} e^{-0.2 x} \hat{a}_y\), find the power required to move the conductor at a constant speed to r = 2.0m, y=0 in 5 x 10-3s. Assume parallel motion along the x-axis. Here \(\hat{a}_x, \hat{a}_y, \hat{a}_z\) are unit vectors along the x, y, and z axes, respectively.

1. 1.57 W
2. 2.97 W
3. 14.85 W
4. 29.7 W

Answer: 2. 2.97 W

Work, \(W=\int_0^2 F d x\)

F = BIl

⇒ \(3 \times 10^{-4} e^{-0.2 x} \times 10 \times 3[∵ l=1.5+1.5=3 \mathrm{~m}]\)

∴ \(W=\int_0^2 3 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x\)

⇒ \(9 \times 10^{-3} \int_0^2 e^{-0.2 x} d x\)

⇒ \(=\frac{9 \times 10^{-3}}{0.2}\left(-e^{-0.2 \times 2}+1\right)=\frac{2.97 \times 10^{-3}}{0.2} \mathrm{~J}\)

Power, \(P=\frac{W}{t}=\frac{2.97 \times 10^{-3}}{0.2 \times 5 \times 10^{-3}}=2.97 \mathrm{~W}\)

The option 2 is correct.

Question 72. Two long current-carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle 0 with the vertical. If the wires have mass A, per unit length then the value of I is (g = gravitational acceleration)

1. \(\sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\)
2. \(2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\)
3. \(2 \sqrt{\frac{\pi g L}{\mu_0} \tan \theta}\)
4. \(\sqrt{\frac{\pi \lambda g L}{\mu_0} \tan \theta}\)

Answer: 2. \(2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\)

The current I is flowing in opposite directions in the two wires.

Therefore, the repulsive force per unit length of the wires,

⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I^2}{2 x}=\frac{\mu_0}{4 \pi} \frac{P^2}{L \sin \theta} \quad[x=L \sin \theta]\)

Weight of the wire per unit length = λg.

If T is the tension in the string, in equilibrium condition,

Tsinθ = F and Tcosθ = λg

⇒ \(\tan \theta=\frac{F}{\lambda g}=\frac{\mu_0 I^2}{4 \pi \lambda g L \sin \theta}\)

⇒ \(I=\sqrt{\frac{4 \pi \lambda g L \sin ^2 \theta}{\mu_0 \cos \theta}}=2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\) \(\left[∵ \sin \theta \tan \theta=\frac{\sin ^2 \theta}{\cos \theta}\right]\)

The option 2 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 73. Two coaxial solenoids of different radii carry current I in the same direction. Let \(\vec{F}_1\) be the magnetic force on the inner solenoid due to the outer one and \(\vec{F}_2\) be the magnetic force on the outer solenoid due to the inner one. Then

1. \(\vec{F}_1=\vec{F}_2=0\)
2. \(\vec{F}_1 \text { is radially inwards and } \vec{F}_2 \text { is radially outwards }\)
3. \(\vec{F}_1 \text { is radially inwards and } \vec{F}_2=0\)
4. \(\overrightarrow{F_1} \text { is radially outwards and } \overrightarrow{F_2}=0\)

Answer: 1. \(\vec{F}_1=\vec{F}_2=0\)

The two coaxial solenoids are actually two magnets of the same strength, the inner solenoid being completely surrounded by the outer solenoid. No resultant force due to a magnet acts on another magnet.

Hence, \(\vec{F}_1=\vec{F}_2=0\)

The option 1 is correct.

Question 74. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below

1. Stable equilibrium
2. Unstable equilibrium

If there is a uniform magnetic field of 00.3 T in the positive z direction, in which orientations the loop would be in

1. (1) and (2), respectively
2. (1) and (3), respectively
3. (2) and (4), respectively
4. (2) and (3), respectively

Answer: 3. (2) and (4), respectively

In cases (2) and (4), torque acting on the loop due to B = 0.

In case (2), if the rectangular loop is slightly disturbed from the xy-plane, the torque acts in such a direction that it tries to bring the loop back to the xy-plane. On the other hand, in case (4), for a similar disturbance of the loop, the acting torque tends to take the loop away from the xy-plane.

It can be noted that in all the cases from (2) to (4), the resultant force acting on the loop = 0.

The option 3 is correct

Question 75. Two identical wires A and B, each of length l, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side a. If BA and BB are the values of the magnetic field at the centers of the circle and square respectively, then the ratio B_A/B_B is

1. \(\frac{\pi^2}{8}\)
2. \(\frac{\pi^2}{16 \sqrt{2}}\)
3. \(\frac{\pi^2}{16}\)
4. \(\frac{\pi^2}{8 \sqrt{2}}\)

Answer: 4. \(\frac{\pi^2}{8 \sqrt{2}}\)

We have, \(l=2 \pi R \quad \text { or, } R=\frac{l}{2 \pi}\)

Again, \(l=4 a \quad \text { or, } a=\frac{l}{4}\)

The magnetic field at the center of the circle,

⇒ \(B_A=\frac{\mu_0 I}{2 R}=\frac{\mu_0 \pi I}{l}\)

The magnetic field at the center of the square,

⇒ \(B_B=4 \times \frac{\mu_0 I}{4 \pi \times \frac{a}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\)

⇒ \(\frac{\mu_0 I}{\frac{\pi l}{8}} \times \frac{2}{\sqrt{2}}=\frac{16 \mu_0 I}{\pi \sqrt{2} l}\)

∴ \(\frac{B_A}{B_B}=\frac{\mu_0 \pi I}{l} \times \frac{\pi \sqrt{2} l}{16 \mu_0 I}=\frac{\pi^2}{8 \sqrt{2}}\)

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 76. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0- 10 V is

1. 1.985 x 10³Ω
2. 2.045 x 10³Ω
3. 2.535 x 10³Ω
4. 4.005 x 10³Ω

Answer: 1. 1.985 x 10³Ω

Let the value of the resistance be put in series by R.

∴ V= i_g(R_G+R)

or, \(R=\frac{V}{i_g}-R_G=\frac{10}{5 \times 10^{-3}}-15\)

= 1.985 x 10³Ω

The option 1 is correct.

Question 77. An electron, a proton, and an alpha particle having the same kinetic energy are moving in circular orbits of radii r_e, r_p, and r_α respectively in a uniform magnetic field B. The relation between r_g, r_p, r_α is

1. \(r_e<r_p<r_\alpha\)
2. \(r_e<r_\alpha<r_p\)
3. \(r_e>r_p, r_e>r_\alpha\)
4. \(r_e<r_p, r_e<r_\alpha\)

Answer: 4. \(r_e<r_p, r_e<r_\alpha\)

For a charged particle moving in a circular orbit,

⇒ \(r=\frac{\sqrt{2 m E}}{q B}\) [ r = radius of the circular orbit, E = kinetic energy of the particle]

∴ \(\frac{r_a}{r_p}=\frac{\sqrt{2 m_a}}{q_a} \times \frac{q_p}{\sqrt{2 m_p}}=1\) \(\text { since } m_\alpha=4 m_p \text { and } q_\alpha=2 q_p\)

∴ r_α = r_p

Now, since the mass of an electron is less than the mass of both proton and a -particle, r_e will be less than both r_p and r_α.

r_e< r_p , r_e < r_α

The option 4 is correct.

Question 78. Two identical long conducting wires AOB and COD are placed at right angles to each other, with one above the other such that 0 is their common point for two. The wires carry I₁ and I₂ currents respectively. A point P is at a height d above the point 0, with respect to the plane of the wires. The magnetic field at P is,

1. \(\frac{\mu_0}{2 \pi d}\left(\frac{I_1}{I_2}\right)\)
2. \(\frac{\mu_0}{2 \pi d}\left(I_1+I_2\right)\)
3. \(\frac{\mu_0}{2 \pi d}\left(I_1^2-I_2^2\right)\)
4. \(\frac{\mu_0}{2 \pi d}\left(I_1^2+I_2^2\right)^{1 / 2}\)

Answer: 4. \(\frac{\mu_0}{2 \pi d}\left(I_1^2+I_2^2\right)^{1 / 2}\)

The magnetic field at P due to current I₁,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \frac{2 I_1}{d}\)

The magnetic field at P due to current I₂,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \frac{2 I_2}{d}\)

∴ The resultant magnetic field due to AOB and COD,

⇒ \(B=\sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2 I_1}{d}\right)^2+\left(\frac{\mu_0}{4 \pi} \frac{2 I_2}{d}\right)^2}\)

∴ \(B=\frac{2 \mu_0}{4 \pi d} \sqrt{I_1^2+I_2^2}\)

⇒ \(\frac{\mu_0}{2 \pi d}\left(I_1^2+I_2^2\right)^{\frac{1}{2}}\)

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 79. A wire carrying a current I has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to the x-axis while a semicircular portion of radius R is lying on the yz-plane. The magnetic field at point O is

1. \(\vec{B}=\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)
2. \(\vec{B}=-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}-2 \hat{k})\)
3. \(\vec{B}=-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)
4. \(\vec{B}=\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}-2 \hat{k})\)

Answer: 3. \(\vec{B}=-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)

The magnetic fields of the linear parts of the wire are directed downwards, i.e., along the negative z-direction; the value of the magnetic field in each case

⇒ \(\frac{\mu_0}{4} \frac{I}{R}\)

Again, the magnetic field for the semicircular portion of the wire at point 0 is directed along the negative x-axis and its value

⇒ \(\frac{1}{2} \frac{\mu_0 I}{2 R}=\frac{\mu_0 I}{4 R}\)

∴ Resultant magnetic field

⇒ \(-\hat{k}^{\mu_0} \frac{I}{4 \pi} \frac{I}{R}-\hat{k}^{\mu_0} \frac{I}{4 \pi} \frac{I}{R}-\hat{i} \frac{\mu_0 I}{4 R}\)

⇒ \(-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)

The option 3 is correct.

Question 80. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the center has a magnitude

1. \(\frac{\mu_0 n e}{2 \pi r}\)
2. Zero
3. \(\frac{\mu_0 n^2 e}{r}\)
4. \(\frac{\mu_0 n e}{2 r}\)

Answer: 4. \(\frac{\mu_0 n e}{2 r}\)

Equivalent current, \(I=\frac{e}{T}=\frac{e}{\frac{1}{n}}=e n\)

Hence, \(B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 n e}{2 r}\)

The option 4 is correct.

Question 81. A square loop ABCD carrying a current i Is placed near and coplanar with a long straight conductor XY carrying a current I. The net force on the loop will be

1. \(\frac{\mu_0 I i}{2 \pi}\)
2. \(\frac{2 \mu_0 I i L}{3 \pi}\)
3. \(\frac{\mu_0 I i L}{2 \pi}\)
4. \(\frac{2 \pi_0 I i}{3 \pi}\)

Answer: 4. \(\frac{2 \pi_0 I i}{3 \pi}\)

The force of attraction on part AB of the loop by conductor XY

⇒ \(\frac{\mu_0 I i L}{2 \pi \times \frac{L}{2}}=\frac{\mu_0 I i}{\pi}\)

Similarly, the repulsive force on part CD of the loop by conductor XY

⇒ \(\frac{\mu_0 I i L}{2 \pi \times 3 \frac{L}{2}}=\frac{\mu_0 I i}{3 \pi}\)

The magnetic force on AD and BC by the conductor XY will be zero. Therefore, the net attractive force on the loop

⇒ \(A B C D=\frac{\mu_0 I i}{\pi}-\frac{\mu_0 I i}{3 \pi}=\frac{2 \pi_0 I i}{3 \pi}\)

The option 4 is correct

Question 82. 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 a radial distance \(\frac{a}{2}\) and 2a respectively, from the axis of the wire is

1. \(\frac{1}{2}\)
2. 1
3. 4
4. \(\frac{1}{4}\)

Answer: 2. 1

Magnetic field B at radial distance \(\frac{a}{2}\) from the axis of the wire can be calculated using Ampere’s circuital law as follows,

⇒ \(B \times 2 \pi \frac{a}{2}=\mu_0 \frac{I}{\pi a^2} \times \pi\left(\frac{a}{2}\right)^2 \quad \text { or, } B=\frac{\mu_0 I}{4 \pi a}\)

Similarly, the magnetic field B’ at radial distance 2a is given by,

⇒ \(B^{\prime} \times 2 \pi \times 2 a=\mu_0 I \quad \text { or, } B^{\prime}=\frac{\mu_0 I}{4 \pi a} \quad ∴ \frac{B}{B^{\prime}}=1\)

The option 2 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 83. An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The space between the plates is filled with the uniform magnetic field of intensity B, as shown in the figure. Neglecting the effect of gravity, the time of straight-line motion of the electron in the capacitor is

1. \(\frac{\epsilon_0 l B}{\sigma}\)
2. \(\frac{\sigma}{\epsilon_0 l B}\)
3. \(\frac{\epsilon_0 B}{\sigma}\)
4. \(\frac{\sigma}{\epsilon_0 B}\)

Answer: 1. \(\frac{\epsilon_0 l B}{\sigma}\)

Electric field between the capacitor, \(E=\frac{\sigma}{\epsilon_0}\)

Force upon the electron = \(-e(\vec{E}+\vec{v} \times \vec{B})\)

For the straight-line motion of the electron, this force must be zero.

∴ \(\vec{E}=-\vec{v} \times \vec{B}\)

Taking only the magnitudes,

⇒ \(E=\nu B \text { or, } v=\frac{E}{B}=\frac{\sigma}{\epsilon_0 B}\)

∴ Time taken by the electron, \(t=\frac{l}{v}=\frac{\epsilon_0 l B}{\sigma}\)

The option 1 is correct.

Conceptual Questions on Ampere’s Law and Faraday’s Law

Question 84. A uniform magnetic field of 0.3 T is established along the positive Z-direction. A rectangular loop in XY-plane of sides 10 cm and 5 cm carries a current of I = 12A as shown. The torque on the loop is

1. \(+1.8 \times 10^{-2} \hat{i} \mathrm{~N} \cdot \mathrm{m}\)
2. \(-1.8 \times 10^{-2} \hat{j} \mathrm{~N} \cdot \mathrm{m}\)
3. Zero
4. \(-1.8 \times 10^{-2} \hat{i} \mathrm{~N} \cdot \mathrm{m}\)

Answer: Zero

Torque on a loop, \(\tau=I \vec{A} \times \vec{B}\)

In the given figure the area vector \(\vec{A}\) of the xy-plane is along the positive Z-direction.

also \(\vec{B}\) is directed along the positive Z-direction.

then, \(\vec{A} \times \vec{B}=\overrightarrow{0}\)

The option 3 is correct.

Question 85. A metallic rod of mass per unit length 0.5 kg m^-1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

1. 14.76 A
2. 5.98 A
3. 7.14 A
4. 11.32 A

Answer: 4. 11.32 A

For the equilibrium of the metallic rod,

mgsin30° = Bilcos30°

⇒ \(i=\frac{m g}{B l} \tan 30^{\circ}=\left(\frac{m}{l}\right) \frac{g}{B} \tan 30^{\circ}\)

⇒ \(=0.5 \times \frac{9.8}{0.25} \times \tan 30^{\circ}\)

= 11.32 A

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 86. Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

1. 250Ω
2. 25Ω
3. 40Ω
4. 500Ω

Answer: 1. 250Ω

The current sensitivity of the galvanometer,

⇒ \(\frac{\theta}{I}=I_s=\frac{B N A}{c}\)

Voltage sensitivity of the galvanometer,

⇒ \(\frac{\theta}{V}=V_s=\frac{B N A}{c R_G}\)

or, \(V_s=\left(\frac{B N A}{c}\right) \times \frac{1}{R_G}=I_s \times \frac{1}{R_G}\)

or, \(R_G=\frac{I_s}{V_s}=\frac{5}{20 \times 10^{-3}}=250 \Omega\)

The option 1 is correct.