The pole strength of a bar magnet is $48$ $ampere-metre$ and the distance between its poles is $ 25 \,cm$ . The moment of the couple by which it can be placed at an angle of $30°$ with the uniform magnetic intensity of flux density $0.15 $ $Newton /ampere-metre$ will be.......$Newton × metre$
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(c)$\tau = MB\sin \theta = 48 \times 25 \times {10^{ - 2}} \times 0.15 \times \frac{1}{2}$$ = 0.9\;N \times m$
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