A square loop of edge length 2 m carrying current of 2 A is placed with its edges parallel to the x-y axis. A magnetic field is passing through the x-y plane and expressed as B = B₀(1 + 4x)k, where B₀ = 5 T. The net magnetic force experienced by the loop is _________ N. Numerical answer type. Work it out, then open the answer.

Answer: 160. B varies only with x, so forces on the two sides parallel to x cancel. Sides parallel to y (length a) at x₁ and x₁ + a get opposite forces: Fnet = I a [B(x₁+a) - B(x₁)] B(x₁+a) - B(x₁) = 4B₀ a = 4×5×2 = 40 T Fnet = 2×2×40 = 160 N

A square loop of edge length 2 m carrying current of 2 A is placed… | JEE Main 2024 (9 Apr, Shift 1)

Q18JEE Main 2024 · 9 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

A square loop of edge length $2\text{ m}$ carrying current of $2\text{ A}$ is placed with its edges parallel to the $x\text{-}y$ axis. A magnetic field is passing through the $x\text{-}y$ plane and expressed as $\vec{B} = B_0(1 + 4x)\hat{k}$ , where $B_0 = 5\text{ T}$ . The net magnetic force experienced by the loop is _________ $\text{N}$ .

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

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Answer: 160

$B$ varies only with $x$ , so forces on the two sides parallel to $x$ cancel.
Sides parallel to $y$ (length $a$ ) at $x_1$ and $x_1 + a$ get opposite forces: $F_{net} = I a\,[B(x_1+a) - B(x_1)]$
$B(x_1+a) - B(x_1) = 4B_0 a = 4\times5\times2 = 40\text{ T}$
$F_{net} = 2\times2\times40 = 160\text{ N}$

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