Question:

Two long straight parallel wires, carrying (adjustable) currents \(I_1\) and \(I_2\), 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 \(I_1 I_2\), would be :

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Parallel currents attract (F negative) and antiparallel currents repel (F positive), while the magnitude is proportional to \(|I_1I_2|\); so F is a straight line through the origin with negative slope.
Updated On: Nov 11, 2026
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The Correct Option is D

Solution and Explanation

The force per unit length between two long parallel wires is \(|F| = \frac{\mu_0 I_1 I_2}{2\pi d}\), proportional to the magnitude of \(I_1 I_2\).
If the currents are in the same direction, \(I_1 I_2 \gt 0\) and the wires attract, so by the given convention \(F \lt 0\).
If the currents are in opposite directions, \(I_1 I_2 \lt 0\) and the wires repel, so \(F \gt 0\).
Hence \(F = -\frac{\mu_0}{2\pi d} I_1 I_2\): a straight line through the origin with a negative slope (second and fourth quadrants).
This is graph (4).
Answer: (4) straight line through the origin falling from the upper-left to the lower-right
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