Two isolated north poles of pole strength $16 Am$ and $4 Am$ respectively are at $30 cm$ apart in air. The distance of the neutral point from the stronger pole is (in $cm$ )

10

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20

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15

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25

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Solution

The correct option is B $20$
Given, $m_{1} = 16 Am; m_{2} = 4 Am; d = 30 cm$

Neutral from the weaker pole is,

$x = \frac{d}{\sqrt{\frac{m_{2}}{m_{1}}} + 1} = \frac{30}{\sqrt{\frac{16}{4}} + 1}$

$∴ m = \frac{30}{3} = 10 cm$

The distance of neutral point from the stronger pole is,

$x^{^{'}} = 30 - 10 = 20 cm$

 Hence, $(B)$ is the correct answer.

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