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MHT CET 2024 (11 May, Shift 2) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from MHT CET 2024 (11 May, Shift 2)

Q31MHT CET 2024 · 11 MAY · SHIFT 2Medium

Magnetic Field of Straight Wires

An infinitely long straight conductor carrying current 'I' is bent in a shape as shown in figure. The radius of the circular part of loop is 'r'. The magnetic induction at the centre 'C' is (μ=\mu = permeability of free space)

Figure

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Answer: (c)

  1. NP is semi-infinite ending directly above C at distance rr: μ04πIr\frac{\mu_0}{4\pi}\frac{I}{r}, out of the page.
  2. Semicircular arc PQ (anticlockwise): μ0I4r=μ04πIrπ\frac{\mu_0 I}{4r} = \frac{\mu_0}{4\pi}\frac{I}{r}\pi, out of the page. The radial part to C gives zero.
  3. B=μ04πIr(1+π)B = \frac{\mu_0}{4\pi}\frac{I}{r}(1 + \pi)
  4. So the answer is (c).
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