A very long magnet of pole strength 16 A-m is placed vertically with its one pole on the table. At what distance from the pole, there will be a neutral point on the table. ( $B_{H}$ = 4 × 10^–5Wbm^–2)

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0.8 m

0.4 m

0.2 m

0.5 m

answer is B.

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Detailed Solution

$\begin{array}{l} B_{H} = \frac{μ_{0}}{4 π} \frac{m}{d^{2}}; 4 \times 10^{- 5} = 10^{- 7} \frac{16}{d^{2}} \\ d^{2} = 0.04, \Rightarrow d = 0.2 m \end{array}$

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