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Question: A uniform horizontal magnetic field of 7.5 x 10-2 T is set up at an angle of 30° with the axis of an...

A uniform horizontal magnetic field of 7.5 x 10-2 T is set up at an angle of 30° with the axis of an solenoid and the magnetic moment associated with it is 1.28 J T-1. Then the torque on it is

A

4.8 x 10-2 N m

B

1.6x10-2N m

C

1.13 x 10-2 kg m

D

4.8 x 10-4 N m

Answer

4.8 x 10-2 N m

Explanation

Solution

: Torque, ,

Here, ,

∴τ=1.28×7.5×10−2sin⁡30∘\therefore \tau = 1.28 \times 7.5 \times 10 ^ { - 2 } \sin 30 ^ { \circ }

=1.28×7.5×10−2×12= 1.28 \times 7.5 \times 10 ^ { - 2 } \times \frac { 1 } { 2 }

=0.64×7.5×10−2=4.8×10−2Nm= 0.64 \times 7.5 \times 10 ^ { - 2 } = 4.8 \times 10 ^ { - 2 } \mathrm { Nm }