The force between two similar poles of strengths $10 A m$ and $40 A m$ separated by a distance of $20 c m$ is

2 \times 10^{- 1} N

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4 \times 10^{- 2} N

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10^{- 3} N

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4 \times 10^{- 3} N

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Solution

The correct option is C $10^{- 3} N$
$F = \frac{μ_{0}}{4 π} \frac{m_{1} m_{2}}{r^{2}} = 10^{- 7} \times 10 \times 40 / (0.2)^{2} = 10^{- 3} N$

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The force between two similar poles of strengths 10 Am and 40 Am separated by a distance of 20 cm is

The correct option is C 10−3NF=μ04πm1m2r2=10−7×10×40/(0.2)2=10−3N

The force between two similar poles of strengths $10 A m$ and $40 A m$ separated by a distance of $20 c m$ is

The correct option is C $10^{- 3} N$
$F = \frac{μ_{0}}{4 π} \frac{m_{1} m_{2}}{r^{2}} = 10^{- 7} \times 10 \times 40 / (0.2)^{2} = 10^{- 3} N$