Question

A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.500 T. If the coil carries a current of 25.0 mA, find the magnitude of the maximum possible torque exerted on the coil.

Answer 1

Answer:

The maximum torque in the coil is $4.9\times 10^{-3}\ N-m$.

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

$\tau=NIAB\ \sin\theta$

For maximum torque, $\theta=90^{\circ}$

$\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m$

So, the maximum torque in the coil is $4.9\times 10^{-3}\ N-m$.