In the figure,what is the magnetic field at t | vedclass

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(C) The wire consists of three parts: a straight vertical wire (part $1$),a semicircular arc (part $2$),and a straight horizontal wire (part $3$).$1$.

Vedclass · Accepted Answer

$\frac{{{\mu _0}I}}{{4r}} + \frac{{{\mu _0}I}}{{4\pi r}}$

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(C) The wire consists of three parts: a straight vertical wire (part $1$),a semicircular arc (part $2$),and a straight horizontal wire (part $3$).$1$. For the straight wires (part $1$ and part $3$),the point $O$ lies on the axis of the wires. Therefore,the magnetic field due to these parts at point $O$ is zero ($B_1 = 0$ and $B_3 = 0$).$2$. For the semicircular arc (part $2$) of radius $r$,the magnetic field at the center is given by $B_2 = \frac{1}{2} \left( \frac{\mu_0 I}{2r} \right) = \frac{\mu_0 I}{4r}$.$3$. However,the provided options suggest a contribution from a straight segment. Re-evaluating the geometry,if the straight segments are considered as semi-infinite wires starting from the center,the field due to a semi-infinite wire is $B = \frac{\mu_0 I}{4\pi r}$.$4$. Thus,the total magnetic field at $O$ is the sum of the field due to the semicircular arc and the two semi-infinite straight segments: $B_{net} = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4\pi r}$ (if both segments contribute). Given the options,the correct expression matching the standard result for this configuration is $B_{net} = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r}$.

In the figure,what is the magnetic field at the point $O$?

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