First slide

The bar magnet

Question

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

Moderate

Solution

$\begin{array}{rcl} B_{H} & = & B_{e} = \frac{μ_{0}}{4 π} \frac{m (2 l)}{{(d^{2} + l^{2})}^{3 / 2}} \\ 0.4 x 10^{- 4} & = & \frac{10^{- 7} m (10^{- 1})}{{(225)}^{3 / 2} x 10^{- 6}}; \\ m & = & 13.5 A - m \end{array}$

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