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Q.

A bar magnet of 10cm long is kept with its N-pole pointing north. A neutral point is formed at a distance of 15cm from each pole. Given the horizontal component of earth’s field is 0.4G, the pole strength of the magnet is……. A-m

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answer is 1.35 AM.

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Detailed Solution

2l=10cm l=5cm=5×10-2m from centre of magnet to point=OP=d=152-52cm =200×10-2m BH=μo4πMd2+l232  0.4×10-4=10-7M200×10-22+5×10-2232    we get  M=1.35Am
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