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Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed 20 cm in tan A position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to

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20 cm

$20 x {(2)}^{1 / 3} cm$

$20 x {(2)}^{4 / 3} cm$

$20 x {(2)}^{2 / 3} cm$

answer is B.

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Detailed Solution

$F o r n o d e f l e c t i o n i n \tan A p o s i t i o n \frac{μ}{4 π} (\frac{2 M_{1}}{{r_{1}}^{3}}) = \frac{μ_{0}}{4 π} (\frac{2 M_{2}}{{r_{2}}^{3}})$

$∴ \frac{M_{1}}{M_{2}} = {(\frac{r_{1}}{r_{2}})}^{3}$

or $\frac{1}{2} = {(\frac{20}{r_{2}})}^{3} \Rightarrow r_{2} = 20 {(2)}^{1 / 3} cm$

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