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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 1: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?. 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.
Number of turns on the circular coil, n = 100
Radius of each turn, r = 8.0 cm = 0.08 m
Current flowing in the coil, I = 0.4 A
Magnitude of the magnetic field at the centre of the coil is given by the relation,
| B | = μ0 2πnI / 4π r
Where,
μ0 = Permeability of free space
= 4π × 10–7 T m A–1
| B | = 4π x 10-7 x 2π x 100 x 0.4 / 4π x 0.08
= 3.14 x 10-4 T
Hence, the magnitude of the magnetic field is 3.14 × 10–4 T.
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:
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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 1: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. ....
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Thanks ðŸËÅ
thanks a lot
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Question is incomplete because the value of mean radius is not given.but anyway you use the formula to find magnetic field at the center of coil that B= munot NI divided by 2 multiply mean radius