JEE Main 2015 · 11 Apr · Shift 1 · Physics Q19

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PhysicsSingle correct+4 / -1Question 19 of 90

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 $I_1I_2$ , would be :

An inverted V: F negative on both sides of the origin

Graph of F against I₁I₂ through the origin O: two lines going down from O on both sides (F ≤ 0).

A V-shape: F positive on both sides

Graph of F against I₁I₂ through the origin O: two lines going up from O on both sides (F ≥ 0).

A straight line through the origin with positive slope

Graph of F against I₁I₂ through the origin O: a straight line through O rising to the right.

A straight line through the origin with negative slope

Graph of F against I₁I₂ through the origin O: a straight line through O falling to the right.

Answer and worked solution

Official answer: D

$F=\dfrac{\mu_0I_1I_2}{2\pi d}$ in magnitude. Parallel currents ( $I_1I_2>0$ ) attract, which counts as negative, and antiparallel ones repel. So $F=-\dfrac{\mu_0}{2\pi d}I_1I_2$ : a straight line through the origin with negative slope, graph (4).

Topics:Force on a current-carrying conductor

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 19