A current I is flowing along an infinite,stra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{\left( \frac{\mu_0 I}{6 \pi} \right)^2 + \left( \frac{\mu_0 I}{8 \pi} \right)^2} = \frac{\mu_0 I}{\pi} \sqrt{\frac{1}{36} + \frac{1}{64}} = \frac{\mu_0 I}{\pi} \sqrt{\frac{16 + 9}{576}} = \frac{\mu_0 I}{\pi} \sqrt{\frac{25}{576}} = \frac{5}{24} \left( \frac{\mu_0 I}{\pi} \right)$.

Similar Questions

Give the $SI$ unit of magnetic field from the Biot-Savart law.

Consider the magnetic field produced by a long straight current-carrying wire.

The magnetic field at the centre of a current-carrying circular coil is

Find the magnetic field at point $O$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module