Question

2. A short bar magnet experiences a torque of
magnitude 0.64 J. When it is placed in a
uniform magnetic field of 0.32 T, taking an
angle of 30° with the direction of the field.
The magnetic moment of the magnet is​

Answer 1

Answer:

Torque = MBsin@

0.64 = M × 0.32 × 1/2

M = 4

Explanation:

hope the answer will help you

Answer 2

Explanation:

Torque = MBsinx

0.64 = M * 0.30*sin 30°

M = 4 Am²