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

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

TestHub · Accepted Answer

Field due to a bar magnet on equatorial position.

$B = \frac{μ_{0} m (2 l)}{4 π {(r^{2} + l^{2})}^{3 / 2}}$

Given

$2 l = 8 c m$

$r = \frac{6}{2} = 3 c m$

At the null point,

$B = B_{H}$

$\frac{μ_{0}}{4 π} \frac{m (2 l)}{{(r^{2} + l^{2})}^{3 / 2}} = B_{H}$

$(1 0^{- 7}) \frac{m (8 \times 1 0^{- 2})}{{[{(3 \times 1 0^{- 2})}^{2} + {(4 \times 1 0^{- 2})}^{2}]}^{3 / 2}} = 3.2 \times 1 0^{- 5}$

$\frac{(1 0^{- 7}) m (8 \times 1 0^{- 2})}{{(5 \times 1 0^{- 2})}^{3}} = 3.2 \times 1 0^{- 5}$

$m = \frac{(125) (3.2)}{8} \times 1 0^{- 2}$

$m = 0.5 A m$

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