A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by: (a) (μ₀ i)/(4R) pointed away from the page (b) (μ₀ i)/(4R)[1 - 2/π] pointed away from page (c) (μ₀ i)/(4R)[1 - 2/π] pointed into the page (d) (μ₀ i)/(4R) pointed into the page

Answer: (b). Each semi-infinite straight wire ends at the foot of the perpendicular from P: B = (μ₀ i)/(4π R), and both point into the page. Semicircle A to B (anticlockwise about P): B = (μ₀ i)/(4R) out of the page. Net: (μ₀ i)/(4R) - (2μ₀ i)/(4π R) = (μ₀ i)/(4R)[1 - 2/π], out of the page. So the answer is (b).

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Magnetic Field of Straight Wires

A very long conducting wire is bent in a semi-circular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by:

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Answer: (b)

Each semi-infinite straight wire ends at the foot of the perpendicular from P: $B = \frac{\mu_0 i}{4\pi R}$ , and both point into the page.
Semicircle A to B (anticlockwise about P): $B = \frac{\mu_0 i}{4R}$ out of the page.
Net: $\frac{\mu_0 i}{4R} - \frac{2\mu_0 i}{4\pi R} = \frac{\mu_0 i}{4R}\left[1 - \frac{2}{\pi}\right]$ , out of the page.
So the answer is (b).

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