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MHT CET 2025 (22 Apr, Shift 2) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from MHT CET 2025 (22 Apr, Shift 2)

Q10MHT CET 2025 · 22 APR · SHIFT 2Easy

Force on Current-Carrying Conductors

A wire of length LL carries current II along x - axis. A magnetic field B⃗=B0(i^−j^−k^) T\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k})\text{ T} acts on the wire. The magnitude of magnetic force acting on the wire is

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Answer: (d)

  1. F⃗=IL⃗×B⃗\vec{F} = I\vec{L} \times \vec{B} with L⃗=Li^\vec{L} = L\hat{i}.
  2. i^×(i^−j^−k^)=−k^+j^\hat{i} \times (\hat{i} - \hat{j} - \hat{k}) = -\hat{k} + \hat{j}
  3. ∣F⃗∣=ILB01+1=2 ILB0|\vec{F}| = ILB_0\sqrt{1 + 1} = \sqrt{2}\,ILB_0
  4. So the answer is (d).
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