Question

A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.50×10−5T and points into the Earth at an angle of 56.0∘ below a line pointing due north. A 7.80-A clockwise current passes through the coil.
Determine the torque on the coil.

Answer 1

Answer:

The torque in the coil is  4.9 × 10⁻⁵ N.m  

Explanation:

T = NIABsinθ

Where;

T is the  torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁵ T

θ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

$A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2$

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°

T = 4.9 × 10⁻⁵ N.m

Therefore, the torque in the coil is  4.9 × 10⁻⁵ N.m