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Q.

The torque required to keep a magnet of length 20 cm at 30º to a uniform field is 2×10–5N-m. The magnetic force on each pole is

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a

2 × 10–4 N

b

4 × 10–4 N

c

2 × 10–3 N

d

2 × 10–6 N

answer is A.

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Detailed Solution

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τ=MBsin300

⇒2×10-5 N=m×0.20×B×12

⇒mB=F=2×10-4 N

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