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JEE Main 2024 (4 Apr, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2024 (4 Apr, Shift 1)

Q19JEE Main 2024 · 4 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

The magnetic field existing in a region is given by B⃗=0.2(1+2x)k^\vec{B} = 0.2(1 + 2x)\hat{k}. A square loop of edge 50 cm50\text{ cm} carrying 0.5 A current is placed in x-yx\text{-}y plane with its edges parallel to the x-yx\text{-}y axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is _________ mN\text{mN}.

Figure

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

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Answer: 50

  1. Forces on the sides parallel to xx cancel (same BB variation along each).
  2. Fnet=Ia [B(x+a)−B(x)]=Ia×0.2×2aF_{net} = I a\,[B(x+a) - B(x)] = I a\times0.2\times2a
  3. Fnet=0.5×0.5×0.4×0.5=0.05 NF_{net} = 0.5\times0.5\times0.4\times0.5 = 0.05\text{ N}
  4. Fnet=50 mNF_{net} = 50\text{ mN}
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