Question

A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?

Answer 1

Explanation:

Coil has 50 turns, n = 50

Intensity of the magnetic field, B = 0.20 T = 2 $\times$ 10−1 T

Coil radius, r  = 2 $\times$ 10−2 m

Current’s magnitude = 5 A

Torque acting on the coil,

$\tau=n i A B \sin \vartheta$

Here, θ is the angle between the magnetic field and the area vector, and A is the coil’s area.

When θ = 90°, τ is maximum.

τmax = niABsin90°

= 6.28 $\times$ 10−2 N-m

Given, θ = 30°, the angle between the coil’s plane and the magnetic field = 90° − 30° = 60°