# 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

##### Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
##### Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
##### NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
##### NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-