Ashok0210
  • 9

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?

  • 9

Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.1

NCERT Solutions Class 12 Physics Chapter 4 Question 1

Share

1 Answer

  1. 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|= (μ0/4π) 2πnl/r

    Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹

    So,

    |B|= (4π x 10⁻⁷)/4π     x   (  2π  x 100 x 0.4 )/r  =3.14 x 10⁻⁴ T

    Hence, the magnitude of the magnetic field is 3.14 x 10⁻⁴ T.

     

    • 0
Leave an answer

Leave an answer

Browse