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Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 8 April, Evening Shift — Question 11

A current carrying circular loop of radius 2cm2\mathrm{cm} with unit normal n^=k^+i^2\hat{\mathbf{n}} = \frac{\hat{\mathbf{k}} + \hat{\mathbf{i}}}{\sqrt{2}} is placed in a magnetic field B⃗=B0(3i^+2k^)\vec{\mathbf{B}} = \mathbf{B}_0(3\hat{\mathbf{i}} + 2\hat{\mathbf{k}}). If B0=4×10−3T\mathbf{B}_0 = 4\times10^{-3}\mathrm{T} and current I=1002A\mathrm{I} = 100\sqrt{2}\mathrm{A}, the torque experienced by the loop is Wb.A. (π=3.14\pi = 3.14)

  1. Option A:

    16×10−5k^16\times10^{-5}\hat{\mathbf{k}}

  2. Option B:

    5024×10−7k^5024\times10^{-7}\hat{\mathbf{k}}

  3. Option C:

    5024×10−7i^5024\times10^{-7}\hat{\mathbf{i}}

  4. Option D:

    5024×10−7j^5024\times10^{-7}\hat{\mathbf{j}}

    Correct

Answer: D

Step-by-step solution

Magnetic moment M⃗=IAn^=1002×π(0.02)2×i^+k^2=100×3.14×4×10−4(i^+k^)=0.1256(i^+k^)\vec{M} = I A \hat{n} = 100\sqrt{2} \times \pi (0.02)^2 \times \frac{\hat{i}+\hat{k}}{\sqrt{2}} = 100\times3.14\times4\times10^{-4} (\hat{i}+\hat{k}) = 0.1256(\hat{i}+\hat{k}). Then τ⃗=M⃗×B⃗=0.1256(i^+k^)×4×10−3(3i^+2k^)=5.024×10−4[(i^+k^)×(3i^+2k^)]=5.024×10−4(2j^−3j^)=−5.024×10−4j^\vec{\tau} = \vec{M}\times\vec{B} = 0.1256(\hat{i}+\hat{k}) \times 4\times10^{-3}(3\hat{i}+2\hat{k}) = 5.024\times10^{-4} [(\hat{i}+\hat{k})\times(3\hat{i}+2\hat{k})] = 5.024\times10^{-4} (2\hat{j} -3\hat{j}) = -5.024\times10^{-4}\hat{j}. Magnitude 50.24×10−5j^50.24\times10^{-5}\hat{j} which equals 5024×10−7j^5024\times10^{-7}\hat{j}.

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Exam
JEE Main 2026
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Force and Torque on Wires and Loops, Magnetic Dipole Moment

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