Question

The pole strength of a bar magnet is 48 ampere-metre and the distance between its poles is 25 cm. The moment of the couple by which it can be placed at an angle of 30o with the uniform magnetic intensity of flux density 0.15 Newton /ampere-metre will be
A) 12 Newton x metre B) 18 Newton x metre C) 0.9 Newton x metre D) None of the above

Answer 1

The torque will be τ = 0.9 N-m

Option (C) is correct.

Explanation:

We are given that:

• Strength of bar magnet =  48 Am
• Distance = 25 cm = 25 x 10^-2 m
• Angle = 30°
• Flux density = 0.15 N / Am

τ = mBsinθ  ---- (1)

M = m x L

M = 48 x 25 x 10^-2

τ = 48 x 25 x 10^-2 x 0.15

τ = 0.9 N-m

Thus the torque will be τ = 0.9 N-m

Also learn more

A 40 kg boy sits 1.2 m from the fulcrum of a see-saw. Where should another boy of 30 kg sit on  the other side in order to balance it?​

https://brainly.in/question/10615840

Answer 2

Answer:

0.9Nm

Explanation:

pls refer to the attachment

Hope you understood....

All the Best

Keep simling ☺