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Physics Question on Moving charges and magnetism

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to . X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is

A

B→=−μ04πIR(πi^+2k^)\overrightarrow{B}=-\frac{\mu_0}{4 \pi}\frac{I}{R}\Big(\pi \hat{i}+2 \hat{k}\Big)

B

B→=μ04πIR(πi^−2k^)\overrightarrow{B}=\frac{\mu_0}{4 \pi}\frac{I}{R}\Big(\pi \hat{i}-2 \hat{k}\Big)

C

B→=μ04πIR(πi^+2k^)\overrightarrow{B}=\frac{\mu_0}{4 \pi}\frac{I}{R}\Big(\pi \hat{i}+2 \hat{k}\Big)

D

B→=−μ04πIR(πi^−2k^)\overrightarrow{B}=-\frac{\mu_0}{4 \pi}\frac{I}{R}\Big(\pi \hat{i}-2 \hat{k}\Big)

Answer

B→=−μ04πIR(πi^+2k^)\overrightarrow{B}=-\frac{\mu_0}{4 \pi}\frac{I}{R}\Big(\pi \hat{i}+2 \hat{k}\Big)

Explanation

Solution

Given situation is shown in the figure.

Parallel wires 1 and 3 are semi-infinite, so magnetic field at O due to them
B1→=B3→=−μ0I4πRk^\overrightarrow{B_1}=\overrightarrow{B_3}=-\frac{\mu_0 I} {4\pi R}\hat{k}
Magnetic field at O due to semi-circular are in
YZ-plane is given by B2→=−μ0I4Ri^\overrightarrow{B_2}=-\frac{\mu_0 I} {4 R}\hat{i}
Net magnetic field at point O is given by
    B→=B1→+B2→+B3→\, \, \, \, \overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}
=−μ0I4πRk^−μ0I4Ri^−μ0I4πRk^=-\frac{\mu_0 I} {4\pi R}\hat{k}-\frac{\mu_0 I} {4 R}\hat{i}-\frac{\mu_0 I} {4\pi R}\hat{k}
=−μ0I4πR(πi^+2k^)=-\frac{\mu_0 I} {4\pi R}(\pi \hat{i}+2 \hat{k})