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Standard XII

Physics

Magnetic Flux and Faraday's Law

Two short bar...

Question

Two short bar magnets with magnetic moments $8$ $A m^{2}$ and $27$ $A m^{2}$ are placed $35$ $c m$ apart along their common axial line with their like poles facing each other. The neutral point is :

in midway between them

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21 c m

from weaker magnet

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14 c m

from weaker magnet

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27 c m

from weaker magnet

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Solution

The correct option is C $14 c m$ from weaker magnet
$m_{1} = 8 A m^{2}$
$m_{2} = 27 A m^{2}$
$B_{1} = \frac{2 μ_{0} m_{1}}{4 π d^{3}}, B_{2} = \frac{2 μ_{0} m_{2}}{4 π {(35 - d)}^{3}}$
$B_{1} = B_{2}$
$\frac{8}{d^{3}} = \frac{27}{{(35 - d)}^{3}}$
$\frac{2}{d} = \frac{3}{(35 - d)}$
$d = 14$ $c m$ (From the weaker magnet)

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Two short bar magnets with magnetic moments 8 Am2 and 27 Am2 are placed 35 cm apart along their common axial line with their like poles facing each other. The neutral point is :

The correct option is C 14 cm from weaker magnetm1=8Am2m2=27Am2B1=2μ0m14πd3,B2=2μ0m24π(35−d)3B1=B28d3=27(35−d)32d=3(35−d)d=14 cm (From the weaker magnet)

Two short bar magnets with magnetic moments $8$ $A m^{2}$ and $27$ $A m^{2}$ are placed $35$ $c m$ apart along their common axial line with their like poles facing each other. The neutral point is :