Two identical bar magnets with a length 10 cm and weight 50 gm-weight are arranged freely with their like poles facing in a inverted vertical glass tube. The upper magnet hangs in the air above the lower one so that the distance between the nearest pole of the magnet is 3mm. Pole strength of the poles of each magnet will be

6.64 a m p \times m

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2 a m p \times m

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10.25 a m p \times m

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Solution

The correct option is A $6.64 a m p \times m$
The weight of upper magnet should be balanced by the repulsion between the two magnet
$∴$ $\frac{μ}{4 π} . \frac{m^{2}}{r^{2}} = 50 g m - w t$
$\Rightarrow$ $10^{- 7} \times \frac{m^{2}}{(9 \times 10^{- 6})} = 50 \times 10^{- 3} \times 9.8$
$\Rightarrow$ $m = 6.64 a m p \times m$

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The correct option is A 6.64amp×mThe weight of upper magnet should be balanced by the repulsion between the two magnet ∴ μ4π.m2r2=50gm−wt ⇒ 10−7×m2(9×10−6)=50×10−3×9.8 ⇒ m=6.64amp×m