A current carrying circular loop of radius… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

A current carrying circular loop of radius $2$ cm with unit normal $\hat{n} = \dfrac{\hat{k}+\hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\vec{B} = B_0(3\hat{i}+2\hat{k})$ . If $B_0 = 4\times 10^{-3}$ T and current $I=100\sqrt{2}$ A, the torque experienced by the loop is ________ Wb·A. ( $\pi=3.14$ )

Official previous-year question

Held on 8 Apr 2026 · Verified 6 Jul 2026.

Options

A
$16\times 10^{-5}\,\hat{k}$
B
$5024\times 10^{-7}\,\hat{k}$
C
$5024\times 10^{-7}\,\hat{i}$
D
$5024\times 10^{-7}\,\hat{j}$

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Solution

The magnetic dipole moment of the current carrying loop is given by $\vec{M} = I A \hat{n}$ .

Given radius $r = 2$ cm $= 2 \times 10^{-2}$ m, the area of the loop is:

$A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4}$ m $^2$

Substituting the given values $I = 100\sqrt{2}$ A and $\hat{n} = \dfrac{\hat{i}+\hat{k}}{\sqrt{2}}$ :

$\vec{M} = (100\sqrt{2}) \times (4\pi \times 10^{-4}) \times \left(\dfrac{\hat{i}+\hat{k}}{\sqrt{2}}\right)$

$\vec{M} = 4\pi \times 10^{-2} (\hat{i}+\hat{k})$ A $\cdot$ m $^2$

The torque experienced by the loop in the magnetic field is $\vec{\tau} = \vec{M} \times \vec{B}$ .

Given $\vec{B} = 4 \times 10^{-3} (3\hat{i}+2\hat{k})$ T, we have:

$\vec{\tau} = [4\pi \times 10^{-2} (\hat{i}+\hat{k})] \times [4 \times 10^{-3} (3\hat{i}+2\hat{k})]$

$\vec{\tau} = 16\pi \times 10^{-5} [(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k})]$

Evaluating the cross product:

$(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k}) = (\hat{i} \times 3\hat{i}) + (\hat{i} \times 2\hat{k}) + (\hat{k} \times 3\hat{i}) + (\hat{k} \times 2\hat{k})$

$= 0 + 2(-\hat{j}) + 3(\hat{j}) + 0 = \hat{j}$

Therefore, the torque is:

$\vec{\tau} = 16\pi \times 10^{-5} \hat{j}$

Substituting $\pi = 3.14$ :

$\vec{\tau} = 16 \times 3.14 \times 10^{-5} \hat{j} = 50.24 \times 10^{-5} \hat{j} = 5024 \times 10^{-7} \hat{j}$ Wb $\cdot$ A

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