NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter: Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 5: Physics is a subject that demands regular and consistent practice. In order to master the concepts and formulas, students must understand all aspects of the topics. Students must practice numerical questions multiple times to avoid confusion in the examination. Practice is the key here. In this article we have provided the NCERT Class 12 Physics Chapter 5 Magnetism and Matter solutions. These are prepared by the subject experts and will help students in the 2025 exams.

The exercise questions given in the NCERT books are always considered important as not only do they help students to analyse their understanding of the concepts involved in a chapter but also in preparing for the examinations. Students should solve as many questions as they can to strengthen their understanding and ability to answer questions. 

Also Check:

Continue reading to download the NCERT Solutions for Class 12 Physics Chapter 5 from the direct link given in this article.  

NCERT Class 12 Physics Chapter 5 Magnetism and Matter Solutions

Go through the questions and answers from NCERT Class 12 Chapter 5 Magnetism and Matter. 

EXERCISES 

Q. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10-² J. What is the magnitude of magnetic moment of the magnet?

Sol. Given:

θ=30°

B = 0.25 T

T = 4.5x10-2 J

Torque is given by

T = mB.sinθ

Substitution

4.5x10-2 = m.(0.25).(0.5)

Therefore, 

m = 0.36 J / T

Q. A short bar magnet of magnetic moment m = 0.32 JT-¹ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Sol. Given, 

m = 0.32 J / T

B = 0.15 T

The bar is free to rotate in the plane of the field. It will be in a stable position in which the net torque is zero. Since the field is uniform, there will be no net force. The angle between the field and the moment should be either zero or 180 degrees. This implies that it can be parallel or antiparallel to the field.

(a) For stable equilibrium, magnetic moment m should be parallel to applied magnetic field.

For stable equilibrium

Potential energy

     U = -mB

U = -0.32 x 0.15

U = -0.048 J

(b) For unstable equilibrium, magnetic moment should be antiparallel to applied magnetic field. 

For unstable equilibrium

U = mB

= 0.32 x 0.15

= +0.048 J

Q. A closely wound solenoid of 800 turns and area of cross section 2.5 × 10-⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Sol. Given, 

N = 800

A = 2.5x10-⁴ sq. m

I = 3 A

The solenoid acts like a bar magnet along the axis, determined by flow of current. The magnetic moment associated

m = NIA = 0.6 J / T

Q. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Sol. Horizontal magnetic field, B = 0.25 T

θ=30

Torque is experienced by the solenoid given by

Torque = m B sinθ

               = 0.6 x 0.25 x 0.5

            = 0.075 Nm

Q. A bar magnet of magnetic moment 1.5 J T-¹ lies aligned with the direction of a uniform magnetic field of 0.22 T. 

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? 

(b) What is the torque on the magnet in cases (i) and (ii)?

Sol. Given, 

m = 1.5 J / T

B = 0.22 T

The magnet is aligned with the direction of uniform magnetic field, and the net force is zero.

(a) (i) θ at normal to the field direction is 90

The work required is given by U = mB

U = 1.5 x 0.22

Or U = 0.33 J

(ii) Opposite to the field direction  = 180

The work required will be double. Thus it is 0.66 J

(b) Value of torque 

T = m. B. sin θ 

(i) θ=90

T = m B sin90

= 1.5 x 0.22 x 1

= 0.33 Nm

(ii) θ=180

T = MB sin180

= zero N

Q. A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10-⁴ m² , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid? 

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10-² T is set up at an angle of 30° with the axis of the solenoid?

Sol. Given, 

N = 2000

A = 1.6 x 10-⁴ m² 

I = 4 A

(a) Magnetic moment

m = nAI

m = 1.28 Am²

(b) In uniform magnetic field, the net force is always zero, irrespective of the

magnitude. 

Torque on the solenoid is given by

                     T = m B sinθ

Given   m = 1.28 Am²

B = 0.075 T

θ=30

Therefore, 

T = 0.048 Nm

Q. A short bar magnet has a magnetic moment of 0.48 J T-¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet. 

Sol. Given, 

m = 0.48 J / T

r = 10 cm

The magnet is short, indicating that the length of the magnet is insignificant when compared to the distance at which the magnitude and direction of the magnetic field is to be determined.

r≫l

The magnitude of the magnetic field at a distance of 10 cm by magnet will be different for both the cases. The direction in both the cases will be different too. 

(a) On the axis

Magnetic field formula

B = μ02m/4πr³

B = 9.6 x 10-⁵ T

Direction: South to North

(b) On normal bisector

B = μ0m/4πr³

B’ = 4.8 x 10-⁵ T

Direction: North to South

To download these solutions in pdf, refer to the below link:

Also, check

• CBSE 12th Date Sheet 2025 Download PDF
• CBSE Class 12 Physics Study Material for 2025
• CBSE Class 12 Physics Most Repeated Questions 
• CBSE Class 12 Physics Competency-Based Questions With Answer Key 2024-25
• CBSE Class 12 Physics Topper Answer Sheet for 2025 Exams
• CBSE Class 12 Physics 90 Days Study Plan for 2025 Exam
• NCERT Solutions for Class 12 (2024-2025) All Subjects
• NCERT Books for CBSE Class 12 - Latest Edition
• NCERT Rationalised Content for Class 12