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(C) The distances from the wires to point $P$ are $r_1 = 3 \ m$ and $r_2 = 4 \ m$. The distance between the wires is $5 \ m$. Since $3^2 + 4^2 = 5^2$,

Vedclass · Accepted Answer

$\frac{5}{24} \left( \frac{\mu_0 I}{\pi} \right)$

Vedclass · Answer

(C) The distances from the wires to point $P$ are $r_1 = 3 \ m$ and $r_2 = 4 \ m$. The distance between the wires is $5 \ m$. Since $3^2 + 4^2 = 5^2$, the triangle formed by the two wires and point $P$ is a right-angled triangle with the right angle at $P$.The magnetic field due to the first wire at $P$ is $B_1 = \frac{\mu_0 I}{2 \pi r_1} = \frac{\mu_0 I}{2 \pi (3)} = \frac{\mu_0 I}{6 \pi}$.The magnetic field due to the second wire at $P$ is $B_2 = \frac{\mu_0 I}{2 \pi r_2} = \frac{\mu_0 I}{2 \pi (4)} = \frac{\mu_0 I}{8 \pi}$.Since the currents are in opposite directions, the magnetic field vectors $\vec{B}_1$ and $\vec{B}_2$ are perpendicular to each other at point $P$ because the triangle is right-angled at $P$. Thus, the net magnetic field is $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{\left( \frac{\mu_0 I}{6 \pi} \right)^2 + \left( \frac{\mu_0 I}{8 \pi} \right)^2} = \frac{\mu_0 I}{\pi} \sqrt{\frac{1}{36} + \frac{1}{64}} = \frac{\mu_0 I}{\pi} \sqrt{\frac{16 + 9}{576}} = \frac{\mu_0 I}{\pi} \sqrt{\frac{25}{576}} = \frac{5}{24} \left( \frac{\mu_0 I}{\pi} \right)$.

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