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JEE MAIN 2026 · Magnetic Effects of Current and Magnetism

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 39

Updated
Single correct+4 / -1Answer matches NTA’s final key

A current carrying circular loop of radius 2 cm with unit normal n^=k^+i^2\hat n = \dfrac{\hat k + \hat i}{\sqrt2} is placed in a magnetic field, B⃗=B0(3i^+2k^)\vec B = B_0\left(3\hat i + 2\hat k\right). If B0=4×10−3B_0 = 4 \times 10^{-3} T and current I=1002I = 100\sqrt2 A, the torque experienced by the loop is ______ Wb.A. (π=3.14\pi = 3.14)

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Solution

Solution

Magnetic moment. ∣m⃗∣=Iπr2=1002×3.14×(0.02)2|\vec m| = I\pi r^2 = 100\sqrt2 \times 3.14 \times (0.02)^2, along n^\hat n:

m⃗=1002×3.14×4×10−4⋅i^+k^2=0.1256 (i^+k^).\vec m = 100\sqrt2 \times 3.14 \times 4 \times 10^{-4} \cdot \frac{\hat i + \hat k}{\sqrt2} = 0.1256\,(\hat i + \hat k).

Torque. τ⃗=m⃗×B⃗\vec\tau = \vec m \times \vec B, with (i^+k^)×(3i^+2k^)=2(i^×k^)+3(k^×i^)=−2j^+3j^=j^(\hat i + \hat k) \times (3\hat i + 2\hat k) = 2(\hat i \times \hat k) + 3(\hat k \times \hat i) = -2\hat j + 3\hat j = \hat j:

τ⃗=0.1256×4×10−3 j^=5.024×10−4 j^=5024×10−7 j^.\vec\tau = 0.1256 \times 4 \times 10^{-3}\,\hat j = 5.024 \times 10^{-4}\,\hat j = 5024 \times 10^{-7}\,\hat j.

Answer: 5024×10−7 j^5024 \times 10^{-7}\,\hat j.

Answer checked against the official answer key. The solution is our own working.

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