Physics · 2026
JEE Main · 8 April 2026, Shift 2 · Q39
A current carrying circular loop of radius 2 cm with unit normal n =(k + i)/(√ 2) is placed in a magnetic field, B =B_o(3 i +2 k ). If B_o=4 × 10^-3…
A current carrying circular loop of radius $\displaystyle 2$ cm with unit normal $\displaystyle \hat{n}=\frac{\hat{k}+\hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\displaystyle \vec{B}=B_o(3 \hat{i}+2 \hat{k})$. If $\displaystyle B_o=4 \times 10^{-3} \mathrm{~T}$ and current $\displaystyle I=100 \sqrt{2} \mathrm{~A}$, the torque experienced by the loop is $\displaystyle \_\_\_\_$ Wb.A. $\displaystyle (\pi=3.14)$
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 5024 \times 10^{-7} \hat{j}$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.