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  4. Force of attraction between two parallel current-carrying conductors is F newton per meter. Current through each of them is doubled and reversed. New force in N/m between these conductors is

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Q. Force of attraction between two parallel current-carrying conductors is $F$ newton per meter. Current through each of them is doubled and reversed. New force in $N/m$ between these conductors is

COMEDKCOMEDK 2006Moving Charges and Magnetism

Solution:

Force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ is given by
$F= \frac{\mu_0 I_1 I_2}{2 \pi r} $
Force is attractive if currents are in same direction. If current through each of them is doubled and reversed, then
$F' = \frac{\mu_0 2 I_1 2 I_2}{2 \pi r} = 4 F$
As direction of current in each conductor is reversed, so again currents in each flow in the same direction. So, force will be attractive.