Magnetic Effects & Magnetism

JEE Main level, single correct, medium

183 of 276

Question 183 of 276

A current carrying circular loop of radius 22 cm with unit normal n^=k^+i^2\hat{n} = \dfrac{\hat{k}+\hat{i}}{\sqrt{2}} is placed in a magnetic field, B⃗=B0(3i^+2k^)\vec{B} = B_0(3\hat{i}+2\hat{k}). If B0=4×10−3B_0 = 4\times 10^{-3} T and current I=1002I=100\sqrt{2} A, the torque experienced by the loop is ________ Wb·A. (π=3.14\pi=3.14)

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

The magnetic dipole moment of the current carrying loop is given by M⃗=IAn^\vec{M} = I A \hat{n}.

Given radius r=2r = 2 cm =2×10−2= 2 \times 10^{-2} m, the area of the loop is:
A=πr2=π(2×10−2)2=4π×10−4A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4} m2^2

Substituting the given values I=1002I = 100\sqrt{2} A and n^=i^+k^2\hat{n} = \dfrac{\hat{i}+\hat{k}}{\sqrt{2}}:
M⃗=(1002)×(4π×10−4)×(i^+k^2)\vec{M} = (100\sqrt{2}) \times (4\pi \times 10^{-4}) \times \left(\dfrac{\hat{i}+\hat{k}}{\sqrt{2}}\right)
M⃗=4π×10−2(i^+k^)\vec{M} = 4\pi \times 10^{-2} (\hat{i}+\hat{k}) A⋅\cdotm2^2

The torque experienced by the loop in the magnetic field is τ⃗=M⃗×B⃗\vec{\tau} = \vec{M} \times \vec{B}.
Given B⃗=4×10−3(3i^+2k^)\vec{B} = 4 \times 10^{-3} (3\hat{i}+2\hat{k}) T, we have:
τ⃗=[4π×10−2(i^+k^)]×[4×10−3(3i^+2k^)]\vec{\tau} = [4\pi \times 10^{-2} (\hat{i}+\hat{k})] \times [4 \times 10^{-3} (3\hat{i}+2\hat{k})]
τ⃗=16π×10−5[(i^+k^)×(3i^+2k^)]\vec{\tau} = 16\pi \times 10^{-5} [(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k})]

Evaluating the cross product:
(i^+k^)×(3i^+2k^)=(i^×3i^)+(i^×2k^)+(k^×3i^)+(k^×2k^)(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k}) = (\hat{i} \times 3\hat{i}) + (\hat{i} \times 2\hat{k}) + (\hat{k} \times 3\hat{i}) + (\hat{k} \times 2\hat{k})
=0+2(−j^)+3(j^)+0=j^= 0 + 2(-\hat{j}) + 3(\hat{j}) + 0 = \hat{j}

Therefore, the torque is:
τ⃗=16π×10−5j^\vec{\tau} = 16\pi \times 10^{-5} \hat{j}

Substituting π=3.14\pi = 3.14:
τ⃗=16×3.14×10−5j^=50.24×10−5j^=5024×10−7j^\vec{\tau} = 16 \times 3.14 \times 10^{-5} \hat{j} = 50.24 \times 10^{-5} \hat{j} = 5024 \times 10^{-7} \hat{j} Wb⋅\cdotA

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