Question

A circular loop of radius 0.7cm has 520 turns of wire and carries a current of 3.9A. The axis of the loop makes an angle of 57 degrees with a magnetic field of 0.982T. Find the magnitude of the torque on the loop.

Answer 1

Torque,
`\tau=0.1669\ N-m`

Step-by-step explanation:

It is given that,

Radius of the circular loop, r = 0.7 cm = 0.007 m

Number of turns, N = 520

Current in the loop, I = 3.9 A

The axis of the loop makes an angle of 57 degrees with a magnetic field.

Magnetic field, B = 0.982 T

We need to find the magnitude of torque on the loop. It is given by :

`\tau=\mu* B`

`\tau=NIABsin(90-57)`

`\tau=520* 3.9\ A* \pi (0.007\ m)^2* 0.982\ T\ cos(57)`

`\tau=0.1669\ N-m`

`\tau=0.167\ N-m`

So, the magnitude of torque is 0.1669 N-m. Hence, this is the required solution.