A bar magnet 8 c m long is placed in the magnetic meridian with the N pole pointing towards geographical north. Two neutral points separated by a distance of 6 c m are obtained on the equatorial axis of the magnet. If the horizontal component of the earth's field = 3.2 × 1 0 - 5 T then pole strength of the magnet is

Question

A bar magnet 8 c m long is placed in the magnetic meridian with the N pole pointing towards geographical north. Two neutral points separated by a distance of 6 c m are obtained on the equatorial axis of the magnet. If the horizontal component of the earth's field = 3.2 × 1 0 - 5 T then pole strength of the magnet is, 0.5 A m, 1 A m, 0.25 A m, 2 A m

MARKS App · Accepted Answer

Field due to a bar magnet on equatorial position. B = μ 0 m 2 l 4 π r 2 + l 2 3 / 2 Given 2 l = 8 c m r = 6 2 = 3 c m At the null point, B = B H μ 0 4 π m 2 l r 2 + l 2 3 / 2 = B H 1 0 - 7 m 8 × 1 0 - 2 3 × 1 0 - 2 2 + 4 × 1 0 - 2 2 3 / 2 = 3.2 × 1 0 - 5 1 0 - 7 m 8 × 1 0 - 2 5 × 1 0 - 2 3 = 3.2 × 1 0 - 5 m = 125 3.2 8 × 1 0 - 2 m = 0.5 A m

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