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Q.

A short bar magnet placed with its axis at 300 with an external field of 800 G experiences a torque of 0.016 Nm. What is the magnetic moment of the magnet? (b) The bar magnet is replaced by a solenoid of cross - sectional area 2 × 10–4 m2 and 1000 turns, but of the same magnetic moment, then the current flowing through the solenoid is:

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a

0.25 Am2, 3A

b

0.40Am2, 2A

c

0.15 Am2, 5A

d

0.10 Am2, 2A

answer is A.

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Detailed Solution

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τ=MBsin⁡θ;  0.016=M×800×10-4×12 M=NiA=0.016×2800×10-4=0.4 Am2 i=MNA=0.41000×2×10−4=2A 

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