NCERT Exemplar for Class 12 Physics Chapter 4

NCERT exemplar for class 12 Physics Chapter 4 will help students to advance the knowledge related to the topic of Moving Charge and Magnetism. This provided exemplar by SimplyAcad allows students to cover all the areas and sections comprising the chapter 4 of the physics textbook. The exemplar contains a total of 26 important questions prepared by the top subject-matter experts. The NCERT exemplar for class 12 physics chapter 4 allows students to make their own notes and practise them regularly to score better in the 12th board examinations. Students will be extremely benefited from the solution set, they can easily access this NCERT exemplar for class 12 Physics in this article by scrolling below.

Along with this, there are several other NCERT exemplar for class 12 science of all the chapters provided in a detailed manner.

Access the NCERT Exemplar Class 12 Physics Chapter 4 – Moving Charge and Magnetism

1. Give two details about the wire’s composition that was utilised to suspend the coil in a moving coil galvanometer.

Ans. The wire used to suspend the coil in a moving coil galvanometer has the following two characteristics.

• • Non-brittle conductor
  • Restoring small torque should be used per unit of twist

2. Two equal-length wires are bent into two loop shapes. One of the loops is circular, while the other is square in shape. The same current flows through them while they are hung in a constant magnetic field. Which loop will be subject to more torque? Explain.

Ans. We know that a circular loop has a larger area than a square loop, the torque experienced by the circular loop would be greater than that of the square loop because torque is directly proportional to area. A circular loop hence experiences more torque.

3. What would a charged particle travelling in the direction of a constant magnetic field take?

Ans. As there would be no forces acting on the charged particle, its route would be a straight line going in the direction of a uniform magnetic field.

4. Define a radial magnetic field. What is it? How is it measured in a galvanometer with a moving coil?

Ans. The magnetic field in which the coil’s plane constantly faces the magnetic field’s direction is known as a radial magnetic field. The following methods can be used to obtain it.

• Concavely cutting the pole sections as instructed
• Inserting a cylindrical core made of soft iron between the pole pieces

5. A long, straight solenoid carrying current I, on an axis along which an electron is travelling at velocity v. What will the magnetic field of the solenoid’s magnetic field be doing to the electron?

Ans. F = Bqv sin is the formula for zero as a force acting on a charged particle travelling in a magnetic field.

Since v and B are both along the -solenoid’s axis in this instance, the angle between them is 0°. Consequently, F= qvB sin = 0.

6. Would your response alter if the circular coil in (a) were swapped out for a planar coil that encloses the same space but has a different shape? (All other information remains the same.)

Ans. From relation (1), it may be concluded that the coil’s shape has no bearing on the applied torque’s size. The size of the coil will determine this. Therefore, the answer would remain the same if the circular coil in the example above were swapped out for a planar coil that encloses the same space but has an irregular shape.

7. A charged particle enters a region of a powerful and irregular magnetic field that varies from point to point in both strength and direction, and it exits the region via a convoluted track. If it didn’t collide with anything, would its final speed be the same as its starting speed?

Ans. Yes, the charged particle’s end speed would be the same as its starting speed since the magnetic force can only change the direction of a particle’s velocity, not its magnitude.

8. An electron moving from west to east enters a chamber with a consistent electrostatic field that runs from north to south. Give instructions on how to create a uniform magnetic field in that direction to stop the electron from veering off course.

Ans. In a chamber with a homogeneous electric field running north to south, an electron that is travelling from west to east enters. If the electric force exerted on it is equal to and opposite to the magnetic field, the travelling electron is not deflected. Southward would be the direction of the magnetic force. Fleming’s left-hand rule also states that the magnetic field should be applied vertically downward.

9. Define cyclotron. Explain in detail.

Ans. Protons, deuterons, and other charged – particles, among others, are accelerated by cyclotrons.

It operates on the tenet that a charged particle can be accelerated to extremely high energies by repeatedly passing through a moderate electric field while being subjected to a strong magnetic field.

10. In a chamber, a magnetic field is created that changes in strength from location to location but always faces east to west. A charged particle enters the chamber and moves with constant speed in a straight line without being deflected. What can you say about the particle’s starting velocity?

Ans. The particle’s starting velocity may be parallel or antiparallel to the magnetic field. As a result, it follows a straight course without experiencing any field deflection.

Q.11. In an area where a uniform magnetic field B is directed normally to the plane of the paper, an alpha particle and a proton are travelling in that plane. What will the ratio of the radii of the trajectories in the field of two particles with equal linear momenta be?

Ans. The radius of the path can be represented by the following equation.

R=mvBq

R ∝ 1q

RaRp=qpqa=e2e=12

Here Ra is the radii of – particle and Rp is its proton, and their respective charge are qa and qp.

∴ Ra: Rp= 1:2

Hence, the ratio will be 1:2.

Q.12. In the presence of a magnetic field B, calculate the force that will act on a charged particle of charge q travelling at a velocity. Demonstrate that when this force is present,

(a) the particle’s K.E. remains constant.

(b) its instantaneous power amounts to zero.

Ans. We already know that

The magnetic force can be expressed as F= q(vB)

Since the force’s direction is parallel to the plane containing (vB)

F = qvBsin 90°=qvB

Here, the displacement and the force will be perpendicular to each other.

W = FScos

W = FScos90°=0

KE = 0

Hence, the kinetic energy will be constant in this particular condition.

(b) the expression for instantaneous power is

p = Fvcos

When velocity and force are perpendicular to each other.

p = FScos90° = 0

Hence, the instantaneous power will be zero.

Q.13. Answer the following.

(a) What distinguishes a toroid from a solenoid? Create a diagram of the magnetic field lines in the two scenarios and compare them.

(b) How is a given solenoid’s magnetic field made strong?

Ans. (a) The toroid is a hollow circular ring on which a substantial number of tightly wound wire turns are placed, in contrast to the solenoid, which is made up of a long wire twisted in the shape of a helix with the neighbouring turns closely spaced.

(b) A soft iron core is inserted within a particular solenoid to create a powerful magnetic field inside of it. By running more current through it, it becomes stronger.

Q.14. A current I is carried by a spherical coil with a radius of r and N tightly wrapped turns. Create the expressions for the following.

(a) A magnetic field that is present at its centre

(b) The coil’s magnetic moment

Ans. (a) A circular coil with N turns and radius r carrying a current, I, has a magnetic field at its centre that is

B = 02 NIr

(b) The magnetic moment, M = NIA = NIr²

Q.15. A current of 10A travels down a straight wire. A 2.0 cm gap separates an electron travelling at 107 m/s from the wire. Find the force on the electron that is being affected by its velocity approaching the wire.

Ans. We know that the current running through a straight wire is I=10A,

The speed of the electron is v=107 m/s

The distance between the wire and the electron is R=20cm= 210-2 m

The force that is acting on the travelling electron can be written as

F= qVBsin

The expression for magnetic field is

B = 04 21r

If we substitute the given values,

B = 10-7210210-2=10-4T, and it is known to be to the plane

The force that is acting on the electron can be represented by,

F = qVBsin

F = 1.610-16 N

Hence, the force will be F = 1.610-16 N.

Q.16. A wire coil of 100 turns, each with a radius of 8.0 cm, is used to transport a current of 0.40. How strong is the magnetic field B in the centre within the coil?

Ans. We are given that,

Number of turns on the coils, n = 100

Each turn’s radius, r = 8.0 cm = 0.08 m

The current that is flowing within the coil is I = 0.4 A

The relationship could be used to determine the strength of the magnetic field at the coil’s centre.

B = 04 2nlr

Here, the permeability of free space, 0=410-7T m A-1

B = 410-7421000.40.08

B = 3.1410-4 T

Hence, the magnetic field’s magnitude will be 3.1410-4 T.

Q.17. A 35A current is carried along a long, straight wire. How big is field B at a location 20 cm from the wire?

Ans. The following is given.

The current present in the wire is I = 35 A

The distance between the point and the wire, r = 20 cm = 0.2 m

The magnetic field at a given point is

B = 04 2lr

Here 0= Permeability of free space = 410-7 T m A-1

B = 410-723540.2

B = 3.510-5 T

Hence, the magnetic field’s magnitude at a distance of 20 cm from the wire will be 3.510-5 T.

Q.18. A 50 A current travels north to south via a straight wire in the horizontal plane. Give B’s magnitude and direction at a location 2.5 metres east of the wire.

Ans. The following information is given.

The current in the wire, I = 50 A

The point is 2.5 m away from the wire’s east.

The magnitude of the distance between the point and the wire is r = 2.5 m.

The magnetic field at a given point can be represented by the following relation.

B = 02I4r

Here, 0 = Permeability of free space = 410-7 T m A-1

B = 410-725042.5

B = 410-6 T

Q.19. Inside a solenoid, a 3.0 cm wire with a 10 A current is inserted perpendicular to the axis. It is stated that there is a 0.27 T magnetic field inside the solenoid. What is the wire’s magnetic field strength?

Ans. The following information is given,

The wire’s length, I = 3 cm = 0.03 m

The current that flows through the wire, I = 10 A

Magnetic field, B = 0.27 T

The angle between the current and the magnetic field = 90°

The magnetic force that is exerted on the wire can be represented as

F = BIlsin

If we substitute the values, we get,

F = 0.27100.03 sin90°

F = 8.110-2 N

Hence, the magnetic force that will be exerted on the wire will be 8.110-2 N and the force’s direction can be derived using Flemming’s left-hand rule.

Q.20. A tightly wound solenoid measuring 80 cm long has five layers with 400 turns each. The solenoid’s diameter is 1.8 cm. Calculate the magnitude of B inside the solenoid located near its centre if the current carried is 8.0A.

Ans. The following information is given.

The solenoid length is given as I = 80 cm = 0.8 m

Since the solenoid has five layers of windings with a total of 400 turns each.

The total turns of the solenoid, D = 1.8 cm = 0.018 m

The current that is carried by the solenoid, I = 8.0 A

The magnetic field’s magnitude present inside and near the centre of the solenoid can be represented by the following equation.

B = 0NIl

Here, 0= 410-4 TmA-1 will be the permeability of free space.

When we substitute the values given, we get

B = 410-7200080.8

B = 2.51210-2 T

Hence, the magnitude of the magnetic field present inside and at its centre will be 2.51210-2 T.

Q.21. Answer the following.

(a) A 6.0A current-carrying circular coil with 30 turns and an 8.0cm radius is hung vertically in a 1.0T uniform horizontal magnetic field. The field lines are at an angle to the coil’s normal. Determine how much counter-torque will need to be provided to stop the coil from turning.

(b) Would your response alter if the circular coil in (a) were swapped out for a planar coil that encloses the same space but has a different shape? (All other information remains the same.)

Ans. (a) The information given is as follows.

The number of turns that the circular coil does, n = 30

The radius of the coil, r = 8.0 cm = 0.08 m

The coil’s area= r² = (0.08)² = 0.0201m²

The current that is flowing through the coil, I = 6.0 A

The strength of the magnetic field, = B = 1 T

The angle existing between the normal with the coil surface and the field lines is, = 60°

The magnetic field produces a torque in the coil and hence it turns. The relation, which is used to calculate the counter torque used to keep the coil from spinning is

T = nIBA sin ……………………….(1)

T = 30610.0201sin 60°

Hence, the counter torque that will be applied to keep the coil from turning is 3.133 Nm.

(b) From relation (1), it may be concluded that the coil’s shape has no bearing on the applied torque’s size. The size of the coil will determine this. Therefore, the answer would remain the same if the circular coil in the example above were swapped out for a planar coil that encloses the same space but has an irregular shape.

Q.22. A square plane coil with 200 turns and a surface area of 100 cm2 can carry 5A of constant current. It is positioned in a 0.2 T uniform magnetic field that is acting perpendicular to the coil’s plane. Calculate the coil’s torque when the field’s direction and its plane are at an angle of 60 degrees. What direction will the coil be in when it reaches a stable equilibrium?

Ans. A = 100 cm² = 100(10-4)m² = 10-2 m

N = 200 turns, I = 5A, B = 0.2 T

= 90°-60°=30°

= NIAB sin

= (200)(5)(10-2)(0.2)12

sin 30°=12

= 1 Nm

Hence, when the coil and the magnetic field are parallel, the coil’s equilibrium state will be stable.

Q.23. Answer the following.

(a) In a chamber, a magnetic field is created that changes in strength from location to location but always faces east to west. A charged particle enters the chamber and moves with constant speed in a straight line without being deflected. What can you describe about the particle’s starting velocity?

(b) A charged particle enters a region of a powerful and irregular magnetic field that varies from point to point in both strength and direction, and it exits the region via a convoluted track. If it didn’t collide with anything, would its final speed be the same as its starting speed?

(c) An electron moving from west to east enters a chamber with a consistent electrostatic field that runs from north to south. Give instructions on creating a uniform magnetic field in that direction to stop the electron from veering off course.

Ans. (a) The particle’s starting velocity may be parallel or anti-parallel to the magnetic field. As a result, it follows a straight course without experiencing any field deflection.

(b) Yes, the charged particle’s end speed would be the same as its starting speed since the magnetic force can only change the direction of a particle’s velocity, not its magnitude.

(c) In a chamber with a homogeneous electric field running north to south, an electron travelling from west to east enters. If the electric force exerted on it is equal to and opposite to the magnetic field, the travelling electron is not deflected. Southward would be the direction of the magnetic force. Fleming’s left-hand rule also states that the magnetic field should be applied vertically downward.

Q.24. A 300A current flows via the cables connecting an automobile’s battery to its starting motor (for a short time). If the distance between the wires is 1.5 cm and they are 70 cm long, what is the force per unit length? Also, answer whether the force will be attractive or repulsive.

Ans. The information given is as follows.

Current present in both the wires, I = 300A

The distance between the two wires, r = 1.5 cm = 0.0015 m

The two wire’s lengths, I = 70 cm = 0.7 m

We already know that the force that exists between two wires can be denoted by the following relation.

F = 0l22r

Here, the permeability of free space 0 = 410TmA-1

If we substitute the values given, we get

F = 410-7300220.015

F= 1.2 N

There is a repulsive force between the wires since the direction of the current in them is observed to be opposing.

Q.25. A constant magnetic field of 6.5G (1G=10-4 T) is maintained inside a chamber.

A normal-to-the-field moving electron enters the field at a speed of 4.8106 ms-1. Explain why the electron’s route is a circle. Calculate the orbit’s circumference. (e=1.610-19C, me= 9.110-31 kg)

Ans. It is given that that magnetic field strength is B = 6.5 G = 6.510-4 T

The electron’s speed, v = 4.8106 m/s

The charge present on the electron, e = 1.610-19C

The electron’s mass, me= 9.110-31 kg

The angle that the magnetic field and the shot electron make, = 90°

The magnetic force that is exerted in the magnetic field on the electron can be represented in the following way.

F = ecBsin

The electron in motion is given centripetal force by this force. As a result, the electron begins to travel in a circular path with a radius r.

Hence the centripetal force that is exerted on the electron is

Fe = mv2r

In this equilibrium, the magnetic force and the centripetal force that is exerted on the electron will be equal

That is, Fc = F

mv2r = evBsin

r = mvBesin

r = 9.110-314.81066.510-41.610-19sin90°

r = 4.210-2 m

r = 4.2 cm

Hence, the circular orbit’s radius of the electron will be 4.2 cm.

Q.26. Imagine a cylindrical region having a radius of 10.0 cm containing a magnetic field of 1.5 T, which runs parallel to the axis running east to west. This area is traversed by a wire carrying 7.0A of current in the north-to-south direction. What size and direction is the force acting on the wire in the event that,

1. the axis is intersected by the wire?
2. the wire changes its direction from north to south to northeast to northwest direction?
3. the wire, which runs from the north to south direction, is lowered 6.0 cm from the axis?

Ans. (a) The information given is that,

The strength of the magnetic field, B = 1.5T

The radius of the given cylindrical region, r = 10 cm = 0.1 m

The current present in the wire that is travelling through the cylindrical region, I = 7A

If the axis is intersected by the wire, then the cylindrical region’s diameter will be the same as the wire’s length. Hence, I = 2r = 0.2m

The angle formed by the current and the magnetic field = 90°

We already know that the magnetic force that is acting on the wire can be represented by the following equation.

F = BlIsin

F = 1.570.2sin90°

F = 2.1 N

Hence, the force that is acting on the wire that is in a vertically downward direction will be 2.1 N.

(b) The new length of the wire after it is turned to the northeast to northwest direction can be represented as

11 = 1sin

The angle that exists between the current and the magnetic field, = 45°

The force remaining on the wire,

F = BIII sin = BII = 1.570.2

F = 2.1N

Hence, the force amounting to 2.1 N will act on the wire in a vertically downward manner. This will be independent of the angle because 1sin is fixed.

(c) d = 6.0 cm is the distance to which the wire is lowered from the axis.

Imagine the new wire’s length to be l2

l222= 4(d+r) = 4(10+6)=416

l2= 82 = 16 cm = 0.16 m

The magnetic force exerted on the wire will be

F2 = Bll2 = 1.570.16

F=1.68N

Hence, a force that will act on the wire in a vertically downward manner will be 1.68N.

NCERT Exemplar For Class 12 Science

Students must practise these additional questions for their own benefits, the ncert exemplar are curated by the best subject-matter experts to boost your knowledge on the presented topic. Students can easily access the ncert exemplar for class 12 science by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks.

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