JEE Main 2026 · 8 Apr · Shift 2 · Physics Q39

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A current carrying circular loop of radius 2 cm with unit normal $\hat n=\dfrac{\hat k+\hat i}{\sqrt2}$ is placed in a magnetic field, $\vec B=B_o\left(3\hat i+2\hat k\right)$ . If $B_o=4\times10^{-3}$ T and current $I=100\sqrt2$ A, the torque experienced by the loop is ________ Wb.A. ( $\pi=3.14$ )

16\times10^{-5}\,\hat k

5024\times10^{-7}\,\hat k

5024\times10^{-7}\,\hat i

5024\times10^{-7}\,\hat j

Answer and worked solution

Official answer: D

Magnetic moment.
$\vec m=IA\hat n=100\sqrt2\times3.14\times(0.02)^2\times\frac{\hat i+\hat k}{\sqrt2}=0.1256\left(\hat i+\hat k\right).$

Torque. $(\hat i+\hat k)\times(3\hat i+2\hat k)=-2\hat j+3\hat j=\hat j$ , so
$\vec\tau=0.1256\times4\times10^{-3}\,\hat j=5024\times10^{-7}\,\hat j.$

(D)

Topics:Torque on a current loop

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JEE Main 2026 · 8 Apr · Shift 2, Physics question 39