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MHT CET 2026 (16 Apr, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from MHT CET 2026 (16 Apr, Shift 1)

Q1MHT CET 2026 · 16 APR · SHIFT 1Easy

Force on Current-Carrying Conductors

A wire carrying current 'I' along x axis has length 'L' and it is kept in a magnetic field B(i^+2j^−2k^)B(\hat{i} + 2\hat{j} - 2\hat{k}) T. The magnitude of magnetic force acting on the wire is

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Answer: (a)

  1. F⃗=IL⃗×B⃗\vec{F} = I\vec{L} \times \vec{B} with L⃗=Li^\vec{L} = L\hat{i}.
  2. i^×(i^+2j^−2k^)=2k^+2j^\hat{i} \times (\hat{i} + 2\hat{j} - 2\hat{k}) = 2\hat{k} + 2\hat{j}
  3. ∣F⃗∣=ILB22+22=8 ILB|\vec{F}| = ILB\sqrt{2^2 + 2^2} = \sqrt{8}\,ILB
  4. So the answer is (a).
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