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A bar magnet when placed at an angle of 30⁰ to the direction of magnetic field induction of 5×10^–5T, experiences a moment of couple 2.5 × 10^–6N–m. If the length of the magnet is 5cm its pole strength is

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$2 \times 10^{- 2} Am$

$5 \times 10^{- 2} Am$

5A–m

2A–m

answer is C.

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Detailed Solution

$τ = MBsin θ = (m l) Bsin θ$

$\Rightarrow 2.5 \times 10^{- 6} = m \times 0.05 \times 5 \times 10^{- 5} \times \frac{1}{2}$

$\Rightarrow m = 2 Am$

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