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Q.

A magnet of length 10cm and pole strength 0.5Am is placed normal to a uniform magnetic field of strength 5 × 10^–5 weber m^–2. The moment of the couple acting on it is

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a

zero

b

$5 \times 10^{- 5} Nm$

c

$2.5 \times 10^{- 7} N - m$

d

$2.5 \times 10^{- 6} N - m$

answer is A.

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Detailed Solution

$τ = M B s i n θ = (0.5 \times 0.1) \times 5 \times 10^{- 5} \times 1 = 2.5 \times 10^{- 6} N - m$

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