Question

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to the loop is parallel to a unit vector n^=−0.60i^−0.80j^. Calculate the magnitude of the torque on the loop due to the magnetic field.

Answer 1

Answer:

The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$.

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vector$n=-0.60\hat{i}-0.080\hat{j}$

We need to calculate the torque on the loop

Using formula of torque

$\tau=NIAB\sin\theta$

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

$\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}$

$\tau=4.7\times10^{-4}\ N-m$

Hence, The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$.