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(C) The magnetic field at $O$ is the vector sum of the fields due to the various segments of the wire:$1$. Straight vertical wire of length $L$ at $x

Vedclass · Accepted Answer

$\vec{B}=\frac{-\mu_0 I}{L}\left(1+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}$

Vedclass · Answer

(C) The magnetic field at $O$ is the vector sum of the fields due to the various segments of the wire:$1$. Straight vertical wire of length $L$ at $x = -L$: It extends from $y = -L$ to $y = 0$. The perpendicular distance from $O$ is $L$. The angles subtended at $O$ are $0^{\circ}$ and $45^{\circ}$.$B_1 = \frac{\mu_0 I}{4 \pi L} (\sin 45^{\circ} + \sin 0^{\circ}) = \frac{\mu_0 I}{4 \sqrt{2} \pi L}$. Direction is $-\hat{k}$.$2$. Straight horizontal wire of length $L/2$ from $x = -L$ to $x = -L/2$: It lies on the axis passing through $O$, so $B_2 = 0$.$3$. Semi-circular arc of radius $L/2$: $B_3 = \frac{\mu_0 I}{4 (L/2)} = \frac{\mu_0 I}{2 L}$. Direction is $-\hat{k}$.$4$. Straight horizontal wire from $x = L/2$ to $x = L/4$: It lies on the axis passing through $O$, so $B_4 = 0$.$5$. Quarter-circular arc of radius $L/4$: $B_5 = \frac{1}{4} \frac{\mu_0 I}{2 (L/4)} = \frac{\mu_0 I}{2 L}$. Direction is $-\hat{k}$.$6$. Straight vertical wire of length $3L/4$ at $x = 0$: It lies on the axis passing through $O$, so $B_6 = 0$.Summing these: $\vec{B} = -(B_1 + B_3 + B_5) \hat{k} = -\left( \frac{\mu_0 I}{4 \sqrt{2} \pi L} + \frac{\mu_0 I}{2 L} + \frac{\mu_0 I}{2 L} \right) \hat{k} = \frac{-\mu_0 I}{L} \left( 1 + \frac{1}{4 \sqrt{2} \pi} \right) \hat{k}$.Therefore, option $(C)$ is correct.

Which one of the following options represents the magnetic field $\vec{B}$ at the origin $O$ due to the current $I$ flowing in the given wire segments lying on the $xy$ plane?

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