Question

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?

Answer 1

as you know, solenoid behaves as a bar magnet.so,torque on the solenoid, $\bf{|\tau|=MBsin\theta}$

where $\tau$ is torque on the solenoid , M is magnetic moment of bar magnet , B is magnetic field and $\theta$ is the angle between M and B.

A/C to Exercise 5.15

Magnetic moment , M = 0.6 Am²

given, axis of solenoid makes an angle of 30° with the direction of applied field. so, $\theta=30^{\circ}$

magnetic field, B = 0.25T

$now,\tau=0.6\times0.25sin30{\circ}\\=0.075Nm$

hence, 0.075 Nm is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field.

Answer 2

Answer:

Magnetic moment , M = 0.6 Am²

Given:

Axis of solenoid makes an angle of 30° with the direction of applied field.

So, θ =30∘

Magnetic field, B = 0.25T

$\begin{lgathered}now,\tau=0.6\times0.25sin30{\circ}\\=0.075Nm\end{lgathered}$