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(B) The magnetic field due to a semi-infinite straight wire at a point $O$ located at a perpendicular distance $d$ from one end is given by $B = \frac

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$4 	imes 10^{-4} \,T$

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(B) The magnetic field due to a semi-infinite straight wire at a point $O$ located at a perpendicular distance $d$ from one end is given by $B = \frac{\mu_0 i}{4 \pi d}$.Here, the current $i = 10 \,A$ and the perpendicular distance $d = 0.5 \,cm = 0.5 	imes 10^{-2} \,m = 5 	imes 10^{-3} \,m$.For one straight segment, the magnetic field at $O$ is:$B_1 = \frac{(4 \pi 	imes 10^{-7}) 	imes 10}{4 \pi 	imes (0.5 	imes 10^{-2})} = \frac{10^{-6}}{0.5 	imes 10^{-2}} = 2 	imes 10^{-4} \,T$.Since the current flows in opposite directions in the two parallel segments, the magnetic fields produced by both segments at point $O$ are in the same direction (using the right-hand rule).Therefore, the total magnetic field is $B_{total} = B_1 + B_2 = 2 	imes 10^{-4} + 2 	imes 10^{-4} = 4 	imes 10^{-4} \,T$.The contribution from the semicircular end is negligible if we consider the straight wires as semi-infinite, or if the radius is very small compared to the length. Given the options, the result is $4 	imes 10^{-4} \,T$.

For a wire, as shown in the figure, carrying a current of $10 \,A$, find the magnetic induction field at the point $O$. (Given: $\mu_0 = 4 \pi \times 10^{-7} \,H/m$)

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