Question

A eirecular cóil of 100 turna and radius `(10/sqrt pi)` cm carrying current of 5.0 A is suspended vertically in a uniform horizontal magnetic field of 2.0 T. The field makes an angle 30° with the normal to the coil. Calculate:

1. the magnetic dipole moment of the coil, and
2. the magnitude of the counter torque that must be applied to prevent the coil from turning.

Answer 1

Given: Magnetic field (B) = 2.0 T

Current (I) = 5.0 A

Angle (θ) = 30°

r = `10/sqrt pi` cm

= `10/sqrt pi xx 10^-2` m

= `0.1/pi` m

Area of circular coil (A) = πr²

= `pi(0.1/pi)^2`

= `pi * 0.01/pi`

= 0.01 m²

I. Magnetic dipole moment (m) = N I A

= 100 × 5.0 × 0.01

= 5 A.m²

II. Torque on coil (τ) = mB sin θ

= 5 × 2 × sin 30°

= 10 × 0.5

= 5 N.m