The magnitude of magnetic induction at mid point O due to current arrangement as shown in Fig will be (a) (μ₀ I)/(π a) (b) (μ₀ I)/(4π a) (c) (μ₀ I)/(2π a) (d) 0

Answer: (a). AB and ED lie on the line through O, so they give zero field. BC and ET are semi-infinite wires starting at the foot of the perpendicular from O, each at distance a/2: B = (μ₀ I)/(4π (a/2)) = (μ₀ I)/(2π a). Both point out of the page, so they add: B = 2 × (μ₀ I)/(2π a) = (μ₀ I)/(π a) So the answer is (a).

The magnitude of magnetic induction at mid point O due to current… | JEE Main 2023 (29 Jan, Shift 1)

Q44JEE Main 2023 · 29 JAN · SHIFT 1Medium

Magnetic Field of Straight Wires

The magnitude of magnetic induction at mid point $\text{O}$ due to current arrangement as shown in Fig will be

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Answer: (a)

AB and ED lie on the line through O, so they give zero field.
BC and ET are semi-infinite wires starting at the foot of the perpendicular from O, each at distance $\frac{a}{2}$ : $B = \frac{\mu_0 I}{4\pi (a/2)} = \frac{\mu_0 I}{2\pi a}$ .
Both point out of the page, so they add: $B = 2 \times \frac{\mu_0 I}{2\pi a} = \frac{\mu_0 I}{\pi a}$
So the answer is (a).

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