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Question

$(d)$ 

In given arrangement, net magnetic field at point $P$ is vector sum of magnetic fields of wire $A B$, arc $B C$, wire $C D$, wire $E F$

Vedclass · Accepted Answer

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

Vedclass · Answer

$(d)$ 

In given arrangement, net magnetic field at point $P$ is vector sum of magnetic fields of wire $A B$, arc $B C$, wire $C D$, wire $E F$, arc $F G$ and wire $G H$.

So, $\quad B_{	ext {net }}=B_{A B} \odot+B_{B C} \odot+B_{C D} \odot$

$+B_{E F}+B_{F G} \otimes+B_{G H}$

As, $P$ lies on axis of wires $F E$ and $G H$.

So, $\quad B_{F E}=B_{G H}=0$

Also, magnetic fields of equal arcs $B C$ and $F C$ are in opposite directions.

So, $B_{B C} \odot+B_{F G} \otimes=0$

So,

$B_{	ext {net }} =\left(B_{A B}+B_{C D}ight)$

$=\frac{\mu_0 I}{4 \pi r}+\frac{\mu_0 I}{4 \pi r}=\frac{\mu_0 I}{2 \pi r} T$

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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