Question

A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnitude 0.20 T exists parallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?

Answer 1

Let ABCD be the rectangular loop.

Given:
No. of turns of the coil, n = 50
Magnetic field intensity, B = 0.20 T = 2 × 10⁻¹ T
Magnitude of current, I = 5 A
Length of the loop, l = 20 cm = 20 × 10⁻² m,
Breadth of the loop, w = 10 cm = 10 × 10⁻² m,
So, area of the loop, A = lw = 0.02 m²,
fig
(a) There is no force on the sides ab and cd, as they are along the magnetic field.
But the force on the sides ad and bc are equal and opposite; so, they cancel each other.
Hence, the net force on the loop is zero.

(b) Torque acting on the coil,
τ = niABsinθ
Here, A is the area of the coil and θ is the angle between the area vector and magnetic field.
τ = niABsin90°
= 1 × 5 × 0.02 × 0.2
= 0.02 N-m
So, the torque acting on the loop is 0.02 N-m and is parallel to the shorter side.