In the given figure,find the magnetic field a | vedclass

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(C) The total magnetic field at point $O$ is the vector sum of the magnetic fields produced by the three segments: the straight wire $AB$,the circular

Vedclass · Accepted Answer

$\frac{3}{8} \frac{\mu_{0} I}{r} + \frac{\mu_{0} I}{4 \pi r}$

Vedclass · Answer

(C) The total magnetic field at point $O$ is the vector sum of the magnetic fields produced by the three segments: the straight wire $AB$,the circular arc $BC$,and the straight wire $CD$.$1$. For the straight wire $AB$,the point $O$ lies on the axis of the wire. Therefore,the magnetic field due to segment $AB$ is $B_{AB} = 0$.$2$. For the straight wire $CD$,the point $O$ is at a perpendicular distance $r$ from the wire. The wire extends from $C$ to infinity. The magnetic field due to a semi-infinite wire is given by $B_{CD} = \frac{\mu_0 I}{4 \pi r}$.$3$. For the circular arc $BC$,the angle subtended at the center $O$ is $	heta = 270^\circ = \frac{3\pi}{2}$ radians. The magnetic field due to a circular arc is $B_{arc} = \frac{\mu_0 I}{2r} \cdot \frac{	heta}{2\pi} = \frac{\mu_0 I}{2r} \cdot \frac{3\pi/2}{2\pi} = \frac{3\mu_0 I}{8r}$.Since the directions of the magnetic fields from the arc and the semi-infinite wire are the same (into the page),the net magnetic field is:$B_{net} = B_{AB} + B_{arc} + B_{CD} = 0 + \frac{3\mu_0 I}{8r} + \frac{\mu_0 I}{4\pi r} = \frac{3}{8} \frac{\mu_0 I}{r} + \frac{\mu_0 I}{4\pi r}$.

In the given figure,find the magnetic field at point $O$.

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