Two long straight parallel wires, carrying (adjustable) currents I1… | JEE Main 2015 (11 Apr, Shift 1)

Q53JEE Main 2015 · 11 APR · SHIFT 1Easy

Force on Current-Carrying Conductors

Two long straight parallel wires, carrying (adjustable) currents I1 and I2, are kept at a distance d apart. If the force 'F' between the two wires is taken as 'positive' when the wires repel each other and 'negative' when the wires attract each other, the graph showing the dependence of 'F', on the product I1I2, would be:

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Answer: (d)

Magnitude $|F| = \frac{\mu_0 I_1 I_2 l}{2\pi d}$ , proportional to the product $I_1 I_2$ .
$I_1 I_2 > 0$ (same direction): wires attract, so $F$ is negative.
$I_1 I_2 < 0$ (opposite directions): wires repel, so $F$ is positive.
So $F = -k\,I_1 I_2$ : a straight line through the origin with negative slope. The answer is (d).

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