Two short magnets with pole strengths of $900 A m$ and $100 A m$ are placed with their axes in the same vertical line, with similar poles facing each other. Each magnet has a length of $1 c m$ . When separation between the nearer poles is $1 c m$ , the weight of the upper magnet is supported by the repulsive force between the magnets. If $g = 10 m s^{- 2}$ , then the mass of upper magnet is ( $x \times 10^{- 4}$ )

1.6 k g

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2.4 k g

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5.5 k g

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4.8 k g

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Solution

The correct option is D $5.5 k g$
$A = \frac{μ_{o}}{4 π} \frac{m_{1} m_{2}}{r^{2}}$ = Force exerted between two mag. poles

$m_{1} = 900, m_{2} = 100, r = 0.01$ .
Total force of repulsion = $F_{T} = A (1 - \frac{2}{2^{2}} + \frac{1}{3^{2}})$
[near poles are like 1cm apart, repulsion, next nearest 2 pairs atract 2cm apart, the furthest pair repulse3 cm apart]
$F_{T} = M g$
$F_{t} = 10^{- 7} \times 900 \times \frac{100}{(0.01)^{2}} \times (\frac{5.5}{9}) = M g = 10 M$
$5.5 \times 10 = 10 M$
Therefore, $M = 5.5 k g$

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