In the given figure,the net magnetic field at | vedclass

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(B) The magnetic field at $O$ is calculated by considering the contributions from different parts of the wire loop.Part $(1)$ and $(5)$ are straight s

Vedclass · Accepted Answer

$\frac{{{\mu _0}i}}{{3\pi a}}\sqrt {4 + {\pi ^2}} $

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(B) The magnetic field at $O$ is calculated by considering the contributions from different parts of the wire loop.Part $(1)$ and $(5)$ are straight segments pointing towards or away from $O$,so their contribution to the magnetic field at $O$ is $0$.Part $(2)$ is a semi-circular arc of radius $r_1 = a/2$. The magnetic field at the center is $B_2 = \frac{\mu_0 i}{4 r_1} = \frac{\mu_0 i}{4(a/2)} = \frac{\mu_0 i}{2a}$ (directed into the page,$-Z$-axis).Part $(4)$ is a semi-circular arc of radius $r_2 = 3a/2$. The magnetic field at the center is $B_4 = \frac{\mu_0 i}{4 r_2} = \frac{\mu_0 i}{4(3a/2)} = \frac{\mu_0 i}{6a}$ (directed out of the page,$+Z$-axis).The net magnetic field along the $Z$-axis is $B_z = B_2 - B_4 = \frac{\mu_0 i}{2a} - \frac{\mu_0 i}{6a} = \frac{\mu_0 i}{3a}$ (into the page).Parts $(3)$ and $(5)$ are straight segments. Using the Biot-Savart law for a finite wire,the field at $O$ is $B = \frac{\mu_0 i}{4\pi d}(\sin 	heta_1 + \sin 	heta_2)$. For these segments,the distance $d$ from $O$ is $a/2$ and $3a/2$ respectively,and the angles are $\pi/2$ and $0$. Summing these components along the $Y$-axis gives $B_y = \frac{\mu_0 i}{4\pi (a/2)} + \frac{\mu_0 i}{4\pi (3a/2)} = \frac{\mu_0 i}{2\pi a} + \frac{\mu_0 i}{6\pi a} = \frac{4\mu_0 i}{6\pi a} = \frac{2\mu_0 i}{3\pi a}$.The net magnetic field is $B_{net} = \sqrt{B_z^2 + B_y^2} = \sqrt{(\frac{\mu_0 i}{3a})^2 + (\frac{2\mu_0 i}{3\pi a})^2} = \frac{\mu_0 i}{3a} \sqrt{1 + \frac{4}{\pi^2}} = \frac{\mu_0 i}{3\pi a} \sqrt{\pi^2 + 4}$.

In the given figure,the net magnetic field at $O$ will be:

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