Question

A 200-turn circular loop of radius 50.0 cm is vertical, with its axis on an east-west line. A current of 100 A circulates clockwise in the loop when viewed from the east. The Earth’s field here is due north, parallel to the ground, with a strength of 3.00 × 10-5 T. What are the direction and magnitude of the torque on the loop?

Answer 1

Answer:

0.471Nm downwards.

Explanation:

For this problem we need to apply the Torque equation,

That is

$T=NIABsin\theta$

Where N is the number of turns, A is the area, B is magnetic field and \theta is the angle loop mae with B

We calculate the area, so

$A=\pi (50cm)^2(10^2m/cm)^2\\A=0.785m^2$

Replacing in the equation of Torque,

$T=NIABsin\theta$

$t=(200)(100)(0.785)(3*10^{-5})(sin90)$

$t=0.471Nm$

We can conclude that the magnutide of maximum torque is 0.471Nm downwards.