A current I is flowing along an infinite, str | 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

What is the ratio of the magnetic field at point $O$ in the given figures?

$A$ coil of one turn is made of a wire of a certain length,and then from the same length of wire,a coil of two turns is made. If the same current is passed in both cases,what is the ratio of the magnetic inductions at their centers?

The radius of a circular coil is $R$ and it carries a current of $I$ ampere. The intensity of the magnetic field at a distance $x$ on the axis from the centre $(x >> R)$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module