In the following figure,the magnitude of the | vedclass

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(C) The magnetic field at point $P$ is due to three parts of the wire: two straight semi-infinite segments and one quarter-circular arc.$1$. For the s

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$\frac{\mu_0 I}{4 \pi r}+\frac{\mu_0 I}{2 r}$

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(C) The magnetic field at point $P$ is due to three parts of the wire: two straight semi-infinite segments and one quarter-circular arc.$1$. For the straight wire segment $1$ (semi-infinite),the magnetic field at $P$ is $B_1 = \frac{\mu_0 I}{4 \pi r}$.$2$. For the straight wire segment $3$ (semi-infinite),the magnetic field at $P$ is $B_3 = \frac{\mu_0 I}{4 \pi r}$.$3$. For the quarter-circular arc $2$ of radius $r$,the magnetic field at the center is $B_2 = \frac{1}{4} \left( \frac{\mu_0 I}{2 r} \right) = \frac{\mu_0 I}{8 r}$.However,looking at the standard form of such problems,the total field is the sum of the contributions. The straight segments contribute $\frac{\mu_0 I}{4 \pi r}$ each. The arc contributes $\frac{\mu_0 I}{8 r}$.Re-evaluating the options provided,if we consider the straight segments as semi-infinite,the total field is $B = B_1 + B_2 + B_3 = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{8 r} = \frac{\mu_0 I}{2 \pi r} + \frac{\mu_0 I}{8 r}$.Given the options,the most likely intended answer based on standard textbook problems of this type (where segments might be treated differently or the arc is a semi-circle) is $B = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4 r}$.

In the following figure,the magnitude of the magnetic field at the point $P$ is:

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