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(C) The magnetic field due to a long straight wire carrying current $I$ at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For point $P$,the

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$\frac{2}{3} \frac{\mu_0 I}{\pi r}$

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(C) The magnetic field due to a long straight wire carrying current $I$ at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For point $P$,the distance from wire $1$ is $r$. Thus,the magnetic field due to wire $1$ is $B_1 = \frac{\mu_0 I}{2 \pi r}$ (directed into the page).The distance from wire $2$ is $2r + r = 3r$. Thus,the magnetic field due to wire $2$ is $B_2 = \frac{\mu_0 I}{2 \pi (3r)} = \frac{\mu_0 I}{6 \pi r}$ (directed into the page).Since both fields are in the same direction (into the page),the net magnetic field $B_{net}$ is:$B_{net} = B_1 + B_2$$B_{net} = \frac{\mu_0 I}{2 \pi r} + \frac{\mu_0 I}{6 \pi r}$$B_{net} = \frac{\mu_0 I}{2 \pi r} (1 + \frac{1}{3}) = \frac{\mu_0 I}{2 \pi r} (\frac{4}{3}) = \frac{2 \mu_0 I}{3 \pi r}$.

Two very long straight wires are set parallel to each other. Each carries a current $I$ in the same direction and the separation between them is $2r$. The intensity of the magnetic field at point $P$ as shown in the figure is . . . . . . .

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