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JEE Main201910 Apr 2019Morning ShiftPhysicsMagnetic Effects of CurrentActual

Two wires A & B are carrying currents I 1 and I 2 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:

Options

  1. Ax = ± I 1 d I 1 - I 2
  2. Bx = I 1 I 1 + I 2 d and x = I 2 I 1 - I 2 d
  3. Cx = I 2 I 1 + I 2 d and x = I 2 I 1 - I 2 d
  4. Dx = I 1 I 1 - I 2 d and x = I 2 I 1 + I 2 d

Correct answer

A. x = ± I 1 d I 1 - I 2

Step-by-step solution

As current in A and B are opposite in direction, net force on C can never be zero if it is kept between A and B . ∴   μ 0 I 1 I 2 π x = μ 0 I   I 2 2 π x ± d I 1 x ± d = I 2   x I 2 - I 1 x = ±   I 1   d x = ± I 1   d I 2 - I 1

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