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JEE Main 2023 (24 Jan, Shift 2) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2023 (24 Jan, Shift 2)

Q31JEE Main 2023 · 24 JAN · SHIFT 2Medium

Force on Current-Carrying Conductors

A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be x130\frac{x}{130} N. The value of xx is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 9

  1. BB is along the 13 cm side; angle θ\theta between the 5 cm side and the hypotenuse has sin⁡θ=1213\sin\theta = \frac{12}{13}.
  2. F=ILBsin⁡θ=2×0.05×0.75×1213F = I L B \sin\theta = 2\times0.05\times0.75\times\frac{12}{13}
  3. F=0.913=9130 NF = \frac{0.9}{13} = \frac{9}{130}\text{ N}
  4. x=9x = 9
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