# 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?

Given,

Magnetic field strength, B = 0.25 T

Magnetic moment, m = 0.6 JT−1

The angle between the axis of the solenoid and the direction of the applied field,  =  30°.

We know, the torque acting on the solenoid is:

= m x B = mBsinθ
= (0.6 )(0.25 T)(sin 30o

= 0.075 J
= 7.5 x  J

The magnitude of torque is 7.5 x  J.

## Related Chapters

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-