A circular coil of wire consisting of 10 | Class 12 Physics Chapter Moving Charges and Magnetism, Moving Charges and Magnetism NCERT Solutions

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: . 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 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?
Answer:

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,

μ= 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.


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

  • Saniya
  • Feb 15, 2023

Thanks 😊


  • deepto
  • Sep 24, 2020

thanks a lot


  • Ritesh
  • Dec 03, 2019

Thank you bhut hard...👌😁


  • Dipu Dileep
  • Oct 31, 2019

Thanks 😊😊😊😊✌🏼✌🏼✌🏼


  • Kavya
  • Jul 20, 2019

Thank you


  • Saima
  • Mar 01, 2019

Thankyou


  • Ankit kumar
  • Aug 03, 2018

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


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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. ....