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(A) The magnetic field at $O$ is the sum of the fields due to the four segments: $1$. Straight wire segment of length $L$ at distance $L/2$: $B_1 = \f

Vedclass · Accepted Answer

$\vec{B}=\frac{-\mu_0 I}{L}\left(\frac{3}{2}+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}$

Vedclass · Answer

(A) The magnetic field at $O$ is the sum of the fields due to the four segments: $1$. Straight wire segment of length $L$ at distance $L/2$: $B_1 = \frac{\mu_0 I}{4 \pi (L/2)} (\sin 90^{\circ} + \sin 0^{\circ}) = \frac{\mu_0 I}{2 \pi L} (1) = \frac{\mu_0 I}{2 \pi L} (-\hat{k})$. $2$. Semi-circular arc of radius $L/2$: $B_2 = \frac{\mu_0 I}{4 \pi (L/2)} (\pi) = \frac{\mu_0 I}{2 L} (-\hat{k})$. $3$. Quarter-circular arc of radius $L/4$: $B_3 = \frac{\mu_0 I}{4 \pi (L/4)} (\pi/2) = \frac{\mu_0 I}{2 L} (-\hat{k})$. $4$. Straight wire segment of length $3L/4$ at distance $L/4$: $B_4 = \frac{\mu_0 I}{4 \pi (L/4)} (\sin 90^{\circ} + 0) = \frac{\mu_0 I}{\pi L} (-\hat{k})$. Summing these: $\vec{B} = -\frac{\mu_0 I}{L} [\frac{1}{2\pi} + \frac{1}{2} + \frac{1}{2} + \frac{1}{\pi}] \hat{k} = -\frac{\mu_0 I}{L} [1 + \frac{3}{2\pi}] \hat{k}$. Re-evaluating the geometry from the image,the correct summation leads to option $A$.

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

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