Question

A magnetised needle of magnetic moment 3.6×<10-²T-¹ is placed at 30° with the direction of uniform magnetic field of magnitude 2× 10-²T calculate the torque acting on the needle​

Answer 1

Answer:

Given, Magnetic momentum, m=4.8×10−2,θ=300,B=3×10−2T

Therefore, T=mBsinθ

=4.8×10−2×3×10−2×sin300

=4.8×10−2×3×10−2×21

=7.2×10−4J 

Where sin300=21