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

Question: 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.


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Comments

  • Saniya
  • 2023-02-15 12:42:45

Thanks 😊


  • deepto
  • 2020-09-24 17:41:03

thanks a lot


  • Ritesh
  • 2019-12-03 13:57:35

Thank you bhut hard...👌😁


  • Dipu Dileep
  • 2019-10-31 06:15:19

Thanks 😊😊😊😊✌🏼✌🏼✌🏼


  • Kavya
  • 2019-07-20 19:28:33

Thank you


  • Saima
  • 2019-03-01 20:03:44

Thankyou


  • Ankit kumar
  • 2018-08-03 11:54:54

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


Comment(s) on this Question

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. ....