Two magnetic poles each of magnetic pole strength 16 A m are placed at a distance of 2 m apart. Find magnetic force acting on each pole.

4 \times 10^{- 7} N

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16 \times 10^{- 7} N

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32 \times 10^{- 7} N

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64 \times 10^{- 7} N

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Solution

The correct option is C $64 \times 10^{- 7} N$
Given $m_{1} = m_{2} = 16 A m$
$d = 2 m$
$F = \frac{μ_{0}}{4 π} \frac{m_{1} m_{2}}{d^{2}} = \frac{10^{- 7} \times 16 \times 16}{2^{2}}$
$= 64 \times 10^{- 7} N$

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Two magnetic poles each of magnetic pole strength 16 A m are placed at a distance of 2 m apart. Find magnetic force acting on each pole.

The correct option is C 64×10−7NGiven m1=m2=16Amd=2mF=μ04πm1m2d2=10−7×16×1622=64×10−7N

The correct option is C $64 \times 10^{- 7} N$
Given $m_{1} = m_{2} = 16 A m$
$d = 2 m$
$F = \frac{μ_{0}}{4 π} \frac{m_{1} m_{2}}{d^{2}} = \frac{10^{- 7} \times 16 \times 16}{2^{2}}$
$= 64 \times 10^{- 7} N$