Question

Question 5.6: If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-201

Answer 1

$\text{as you know, solenoid behaves as a bar magnet}\\\text{so,torque on the solenoid},\vec{\tau}=\vec{M}\times\vec{B}\\|\tau|=MBsin\theta$
A/C to question,
axis of solenoid makes an angle of 30° with the direction of applied field. so, $\theta=30^{\circ}$
magnetic field, B = 0.25T
from previous question, M = 0.6 Am²
now, $\tau=0.6\times0.25sin30{\circ}\\=0.075Nm$