The ratio of the magnetic field at the centre | vedclass

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

Question

(B) Let the length of the wire be $L$. For a circular loop of radius $R$,$L = 2\pi R$,so $R = \frac{L}{2\pi}$.The magnetic field at the center of the

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the length of the wire be $L$. For a circular loop of radius $R$,$L = 2\pi R$,so $R = \frac{L}{2\pi}$.The magnetic field at the center of the circular loop is $B_1 = \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.For a square coil of side $a$,$L = 4a$,so $a = \frac{L}{4}$.The magnetic field at the center of a square coil is the sum of the fields due to its four sides. For one side,the field is $B_{side} = \frac{\mu_0 i}{4\pi(a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2\pi a} \sqrt{2} = \frac{\mu_0 i}{\sqrt{2}\pi a}$.Total field $B_2 = 4 	imes B_{side} = \frac{4\mu_0 i}{\sqrt{2}\pi a} = \frac{2\sqrt{2}\mu_0 i}{\pi a}$.Substituting $a = L/4$,$B_2 = \frac{2\sqrt{2}\mu_0 i}{\pi (L/4)} = \frac{8\sqrt{2}\mu_0 i}{\pi L}$.The ratio is $\frac{B_1}{B_2} = \frac{\mu_0 i \pi / L}{8\sqrt{2}\mu_0 i / \pi L} = \frac{\pi^2}{8\sqrt{2}}$.

The ratio of the magnetic field at the centre of a current-carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

Similar Questions

$A$ person standing on a rotating platform has his hands lowered. He suddenly outstretches his arms. The angular momentum

The perpendicular distance from the point $(1, \pi)$ to the line joining $(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ (in polar coordinates) is:

The period of the function $f(\theta) = \sin \frac{\theta}{3} + \cos \frac{\theta}{2}$ is (in $\pi$)

If $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has local maximum and minimum at $x = p$ and $x = q$,respectively,then $(p, q)$ equals

The moment of inertia of a sphere (mass $M$ and radius $R$) about its diameter is $I$. Four such spheres are arranged as shown in the figure. The moment of inertia of the system about the axis $XX'$ will be (in $,I$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module