A uniform magnetic field of 3000 G is es | Class 12 Physics Chapter Moving Charges and Magnetism, Moving Charges and Magnetism NCERT Solutions

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Academic session 2026-27, Chapter 4 in the current curriculum.

Welcome to the NCERT Solutions for Class 12 Physics - Chapter Moving Charges and Magnetism. This page offers a step-by-step solution to the specific question from Exercise 1, Question 24:

A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Figure? What is the force on each case? Which case corresponds to stable equilibrium?

. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 24:

A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Figure? What is the force on each case? Which case corresponds to stable equilibrium?

Answer:

Magnetic field strength, B = 3000 G = 3000 × 10-4 T = 0.3 T

Length of the rectangular loop, l = 10 cm

Width of the rectangular loop, b = 5 cm

Area of the loop,

A = l × b = 10 × 5 = 50 cm2 = 50 × 10-4 m2

Current in the loop, I = 12 A

Now, taking the anti-clockwise direction of the current as positive and vise-versa:

(a) Torque, 

From the given figure, it can be observed that A is normal to the y-z plane and B is directed along the z-axis.

The torque 1.8 x 10-2  is  N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.

(b) This case is similar to case (a). Hence, the answer is the same as (a).

(c) Torque 

From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.

The torque 1.8 x 10-2  is  N m along the negative x direction and the force is zero.

(d) Magnitude of torque is given as:

Torque is  1.8 x 10-2 N m at an angle of 240° with positive x direction. The force is zero.

(e) Torque 

Hence, the torque is zero. The force is also zero.

(f) Torque 

Hence, the torque is zero. The force is also zero.

In case (e), the direction of  and  is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.

Whereas, in case (f), the direction of  and  is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Moving Charges and Magnetism.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Comments

  • MD Azeen
  • Sep 26, 2017

How can be 240 degree and f=0 and stability I have not understood.


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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 24: A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop....

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