A current carrying circular loop of radius 2 cm with unit normal n = (k + i)/(√(2)) is placed in a magnetic field, B = Bo(3i + 2k). If Bo = 4 × 10⁻³ T and current I = 100√(2) A, the torque experienced by the loop is ______ Wb.A. ( π = 3.14) (a) 16 × 10⁻⁵k (b) 5024 × 10⁻⁷k (c) 5024 × 10⁻⁷i (d) 5024 × 10⁻⁷j

Answer: (d). m = IA n = 100√2 × π(0.02)² × ( i + k)/(√2) = 0.04π( i + k) = 0.1256( i + k) τ = m × B = 0.1256 × 4×10⁻³ ( i + k)×(3 i + 2 k) ( i + k)×(3 i + 2 k) = -2 j + 3 j = j τ = 5.024×10⁻⁴ j = 5024×10⁻⁷ j So the answer is (d) 5024×10⁻⁷ j.

A current carrying circular loop of radius 2 cm with unit normal n =… | JEE Main 2026 (8 Apr, Shift 2)

Q15JEE Main 2026 · 8 APR · SHIFT 2Medium

Current Loop as a Magnetic Dipole

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

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Answer: (d)

$\vec m = IA\hat n = 100\sqrt2 \times \pi(0.02)^2 \times \frac{\hat i + \hat k}{\sqrt2} = 0.04\pi(\hat i + \hat k) = 0.1256(\hat i + \hat k)$
$\vec\tau = \vec m \times \vec B = 0.1256 \times 4\times10^{-3}\,(\hat i + \hat k)\times(3\hat i + 2\hat k)$
$(\hat i + \hat k)\times(3\hat i + 2\hat k) = -2\hat j + 3\hat j = \hat j$
$\vec\tau = 5.024\times10^{-4}\hat j = 5024\times10^{-7}\hat j$
So the answer is (d) $5024\times10^{-7}\hat j$ .

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