Two short bar magnets with magnetic moments $400 a b A$ ${c m}^{2}$ and $800 a b A$ ${c m}^{2}$ are placed with their axis in the same straight line with similar poles facing each other and with their centres at $20 c m$ apart from each other. Then the force of repulsion is:

Disclaimer: No option as $0.08 d y n e$ was given in the original paper.

12 d y n e

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6 d y n e

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0.08 d y n e

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150 d y n e

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Solution

The correct option is D $0.08 d y n e$
Magnetic force between the poles is:
$F = \frac{μ_{o}}{4 π} \frac{m_{1} m_{2}}{d^{2}}$
Here, $m_{1} = 400 a b A c m^{2}$ $=$ $0.4 A m^{2}$
$m_{2} = 800 a b A c m^{2}$ $=$ $0 .8 A m^{2}$
$d = 20 c m$ .

Substituting these values in the formula:
$F$ = $8 \times 10^{- 7}$ $N$
$F = 0.08 d y n e$ .

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The correct option is D 0.08 dyneMagnetic force between the poles is:F=μo4πm1m2d2Here, m1=400 abA cm2= 0.4 Am2m2=800 abA cm2=0.8 Am2d=20 cm.Substituting these values in the formula:F = 8×10−7 NF=0.08 dyne.