A square frame of side l carries a current i. | vedclass

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

Question

(A) For a square frame of side $l$,the distance from the centre to any side is $d = \frac{l}{2}$.The magnetic field due to one side at the centre is $

Vedclass · Accepted Answer

$\frac{8 \sqrt{2}}{\pi^2}$

Vedclass · Answer

(A) For a square frame of side $l$,the distance from the centre to any side is $d = \frac{l}{2}$.The magnetic field due to one side at the centre is $B_1 = \frac{\mu_0 i}{4 \pi d} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 i}{4 \pi (l/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2 \pi l} \sqrt{2} = \frac{\sqrt{2} \mu_0 i}{2 \pi l}$.Since there are $4$ sides,the total magnetic field at the centre is $B = 4 	imes B_1 = 4 	imes \frac{\sqrt{2} \mu_0 i}{2 \pi l} = \frac{2 \sqrt{2} \mu_0 i}{\pi l}$.For the circular coil,the perimeter is equal to the perimeter of the square,so $2 \pi r = 4l$,which gives $r = \frac{2l}{\pi}$.The magnetic field at the centre of the circular coil is $B^{\prime} = \frac{\mu_0 i}{2r} = \frac{\mu_0 i}{2(2l/\pi)} = \frac{\mu_0 i \pi}{4l}$.Now,the ratio is $\frac{B}{B^{\prime}} = \frac{2 \sqrt{2} \mu_0 i / \pi l}{\mu_0 i \pi / 4l} = \frac{2 \sqrt{2}}{\pi} 	imes \frac{4}{\pi} = \frac{8 \sqrt{2}}{\pi^2}$.

$A$ square frame of side $l$ carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is

Similar Questions

The magnetic field at the centre $C$ of the arrangement shown in the figure is:

$A$ long straight wire of circular cross-section is made of a non-magnetic material. The wire is of radius $a$. The wire carries a current $I$ which is uniformly distributed over its cross-section. The energy stored per unit length in the magnetic field contained within the wire is

$A$ current flows through a circular coil of radius $R$. Find the distance from the center on the axis where the magnetic field is $\frac{1}{8}$th of the magnetic field at the center.

An element $dl = dx \hat{i}$ (where,$dx = 1 \, cm$) is placed at the origin and carries a large current $i = 10 \, A$. What is the magnetic field on the $Y$-axis at a distance of $0.5 \, m$?

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 . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module