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Question: If a solenoid is having magnetic moment of 0.65 J T-1 is free to turn about the vertical direction a...

If a solenoid is having magnetic moment of 0.65 J T-1 is free to turn about the vertical direction and has a uniform horizontal magnetic field of 0.25 T applied. What is the magnitude of the torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

A

0.075 N m

B

0.080 N m

C

0.081 N m

D

0.091 N m

Answer

0.081 N m

Explanation

Solution

: Here,

M−0.65 J T−1, B=0.25 T,θ=30∘\mathrm { M } - 0.65 \mathrm {~J} \mathrm {~T} ^ { - 1 } , \mathrm {~B} = 0.25 \mathrm {~T} , \theta = 30 ^ { \circ }

∴\therefore τ=MBsin⁡θ\tau = \mathrm { MB } \sin \theta

=0.65×0.25×sin⁡30∘= 0.65 \times 0.25 \times \sin 30 ^ { \circ }