Question

What is the maximum torque on a 400-turn circular coil of radius 0.65 cm that carries a current of 1.2 mA and resides in a uniform magnetic field of 0.2 T? Answer in units of N · m.

Answer 1

Answer:

Torque, $\tau=1.27\times 10^{-5}\ N-m$

Explanation:

Given that,

Number of turns in a circular coil, N = 400

Radius of the circular coil, r = 0.65 cm

Current in the coil, I = 1.2 mA

Uniform magnetic field in the circular coil, B = 0.2 T

Torque in the circular coil is given by :

$\tau=NIAB\ \sin\theta$

For maximum torque, $\sin\theta=1$

$\tau=400\times 1.2\times 10^{-3}\times \pi (0.65\times 10^{-2})^2\times 0.2\\\\\tau=1.27\times 10^{-5}\ N-m$

So, the net torque about the vertical axis is $1.27\times 10^{-5}\ N-m$. Hence, this is the required solution.