# 5.8 A closely wound solenoid of 2000 turns and area of cross-section , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.  (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of   is set up at an angle of 30° with the axis of the solenoid?

Now,

Magnetic field strength, B =

The angle between the magnetic field and the axis of the solenoid,

Now, As the Magnetic field is uniform, the Force is zero

Also, we know,

= mxB = mBsinθ

= (1.28 )()(sin )

= 4.8 x  J

Therefore, Force on the solenoid = 0 and torque on the solenoid = 4.8 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/-