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JEE Main 2004 (AIEEE) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2004 (AIEEE)

Q67JEE Main 2004 · AIEEEEasy

Force on Current-Carrying Conductors

Two long conductors, separated by a distance dd carry current I1I_1 and I2I_2 in the same direction. They exert a force FF on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d3d. The new value of the force between them is

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Answer: (a)

  1. F=μ0I1I2l2πd⇒F∝I1I2dF = \frac{\mu_0 I_1 I_2 l}{2\pi d} \Rightarrow F \propto \frac{I_1 I_2}{d}
  2. New: I2→−2I2I_2 \to -2I_2, d→3dd \to 3d
  3. F′=F×(−2)3=−2F3F' = F\times\frac{(-2)}{3} = -\frac{2F}{3} (minus sign: attraction becomes repulsion)
  4. So the answer is (a) −2F3-\frac{2F}{3}.
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