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(D) In the given arrangement,the net magnetic field at point $P$ is the vector sum of the magnetic fields produced by the straight wire segments $AB$,

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) In the given arrangement,the net magnetic field at point $P$ is the vector sum of the magnetic fields produced by the straight wire segments $AB$,$CD$,$EF$,$GH$ and the circular arc segments $BC$ and $FG$.$1$. The point $P$ lies on the axis of the straight wires $EF$ and $GH$. Therefore,the magnetic field due to these segments at $P$ is zero: $B_{EF} = B_{GH} = 0$.$2$. The magnetic field due to a semi-infinite wire at a distance $r$ from its end is $B = \frac{\mu_0 I}{4 \pi r}$.$3$. The magnetic field due to the straight segment $AB$ at $P$ is $B_{AB} = \frac{\mu_0 I}{4 \pi r}$ (directed into the page).$4$. The magnetic field due to the straight segment $CD$ at $P$ is $B_{CD} = \frac{\mu_0 I}{4 \pi r}$ (directed into the page).$5$. The magnetic field due to the quarter-circular arc $BC$ at $P$ is $B_{BC} = \frac{\mu_0 I}{8 r}$ (directed out of the page).$6$. The magnetic field due to the quarter-circular arc $FG$ at $P$ is $B_{FG} = \frac{\mu_0 I}{8 r}$ (directed into the page).$7$. The net magnetic field is $B_{net} = B_{AB} + B_{CD} + B_{FG} - B_{BC} = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{8 r} - \frac{\mu_0 I}{8 r} = \frac{2 \mu_0 I}{4 \pi r} = \frac{\mu_0 I}{2 \pi r}$.Thus,the correct option is $(d)$.

Two infinitely long wires each carrying current $I$ along the same direction are made into the geometry as shown in the figure below. The magnetic field at the point $P$ is

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