In the following figure, the magnitude of the | vedclass

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(C) The magnetic field at point '$O$' is the sum of the magnetic fields due to three segments: the straight wire $AB$, the circular arc $BC$, and the

Vedclass · Accepted Answer

$\frac{\mu_0}{4\pi} \frac{I}{r} (2 + \frac{\pi}{2})$

Vedclass · Answer

(C) The magnetic field at point '$O$' is the sum of the magnetic fields due to three segments: the straight wire $AB$, the circular arc $BC$, and the straight wire $CD$.$1$. For the straight wire $AB$, point '$O$' lies on the line extending from the wire. Therefore, the magnetic field due to $AB$ at '$O$' is $B_1 = 0$.$2$. For the circular arc $BC$ of radius $r$ and angle $	heta = \pi/2$, the magnetic field at the center is given by $B_2 = \frac{\mu_0 I 	heta}{4\pi r} = \frac{\mu_0 I (\pi/2)}{4\pi r} = \frac{\mu_0 I}{8r}$.$3$. For the straight wire $CD$, point '$O$' lies on the line extending from the wire. Therefore, the magnetic field due to $CD$ at '$O$' is $B_3 = 0$.Wait, re-evaluating the geometry: The wire segments $AB$ and $CD$ are semi-infinite wires ending at $B$ and starting at $C$ respectively. The distance from '$O$' to the line of $AB$ is $r$, and from '$O$' to the line of $CD$ is $r$.For a semi-infinite wire, the magnetic field at a perpendicular distance $r$ from the end is $B = \frac{\mu_0 I}{4\pi r}$.- Magnetic field due to $AB$ at '$O$': $B_{AB} = \frac{\mu_0 I}{4\pi r}$.- Magnetic field due to arc $BC$ at '$O$': $B_{arc} = \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} = \frac{\mu_0 I}{8r}$.- Magnetic field due to $CD$ at '$O$': $B_{CD} = \frac{\mu_0 I}{4\pi r}$.Total magnetic field $B = B_{AB} + B_{arc} + B_{CD} = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{8r} + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r} (1 + \frac{\pi}{2} + 1) = \frac{\mu_0 I}{4\pi r} (2 + \frac{\pi}{2})$.Thus, the correct option is $C$.

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

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