Two wires A & B are carrying currents I1 & I2 as shown in the figure… | JEE Main 2019 (10 Apr, Shift 1)

Q47JEE Main 2019 · 10 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

Two wires A & B are carrying currents I1 & I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are:

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Answer: (d)

A and B carry opposite currents. Between them the forces on C from A and B point the same way, so F = 0 is possible only outside the pair.
Zero force needs $\frac{\mu_0 I_1 I}{2\pi |x|} = \frac{\mu_0 I_2 I}{2\pi |x-d|}$ , i.e. $I_1|x-d| = I_2|x|$ with C outside.
If $I_1 > I_2$ , C lies beyond B: $I_1(x-d) = I_2 x \Rightarrow x = \frac{I_1 d}{I_1-I_2}$ . If $I_1 < I_2$ , C lies on the other side of A, at distance $\frac{I_1 d}{I_2-I_1}$ .
Both cases are covered by $x = \pm\frac{I_1 d}{I_1-I_2}$ .
So the answer is (d).

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Force on Current-Carrying Conductors question from JEE Main 2019 (10 Apr, Shift 1)

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