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Two straight parallel wires, both carrying 10 A current in the same direction, attract each other with a force of current $1 \times 10^{- 3} N$ . If both currents are doubled, the force of attraction will be

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$1 \times 10^{- 3} N$

$2 \times 10^{- 3} N$

$4 \times 10^{- 3} N$

$0.25 \times 10^{- 3} N$

answer is C.

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Detailed Solution

$F = \frac{μ_{0}}{4 π} \frac{2 i_{1} i_{2}}{a} \times l = 10^{- 3} N$

When current in both the wires is doubled, then

$F^{'} = \frac{μ_{0}}{4 π} \frac{2 (2 i_{1} ×2 i_{2})}{a} \times l = 4 \times 10^{- 3} N$

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