In the following figure, the magnitude of the magnetic field at point 'O' will be (a) (μ₀)/(4π)I/r(2/π + 2) (b) (μ₀)/(4π)I/r(2/π - 2) (c) (μ₀)/(4π)I/r(2 + π/2) (d) (μ₀)/(4π)I/r(2 - π/2)

Answer: (c). AB and CD are semi-infinite wires ending at the foot of the perpendicular from O: each gives (μ₀)/(4π)I/r. Quarter arc BC: 1/4 · (μ₀ I)/(2r) = (μ₀)/(4π)I/r · π/2 All three are out of the page, so they add: B = (μ₀)/(4π)I/r(2 + π/2) So the answer is (c).

In the following figure, the magnitude of the magnetic field at point… | MHT CET 2026 (18 Apr, Shift 1)

Q10MHT CET 2026 · 18 APR · SHIFT 1Medium

Magnetic Field of Straight Wires

In the following figure, the magnitude of the magnetic field at point 'O' will be

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Answer: (c)

AB and CD are semi-infinite wires ending at the foot of the perpendicular from O: each gives $\frac{\mu_0}{4\pi}\frac{I}{r}$ .
Quarter arc BC: $\frac{1}{4} \cdot \frac{\mu_0 I}{2r} = \frac{\mu_0}{4\pi}\frac{I}{r} \cdot \frac{\pi}{2}$
All three are out of the page, so they add: $B = \frac{\mu_0}{4\pi}\frac{I}{r}\left(2 + \frac{\pi}{2}\right)$
So the answer is (c).

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