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A short bar magnet is placed with its North pole pointing North. Neutral point is at 12 cm from the centre of the magnet. Another short magnet is now placed on the first magnet, with their axes parallel to each other then the neutral point is found to be at 8 cm. The ratio of their magnetic moments is

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3 : 2

27 : 19

9 : 4

9 : 5

answer is B.

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Detailed Solution

$In the first case \frac{μ_{0}}{4 π} \frac{M_{1}}{12^{3}} = B_{H}$ and

$in the second case as the null point becomes closer it is antiparallel$

$\frac{μ_{0}}{4 π} \frac{M_{1} - M_{2}}{8^{3}} = B_{H}$

$\frac{M_{1}}{12^{3}} = \frac{M_{1} - M_{2}}{8^{3}}$

solving $\frac{M_{1}}{M_{2}} = \frac{27}{19}$

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