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 $L$ be the total length of the wire. For the circular coil of radius $R$,$L = 2\pi R$,so $R = \frac{L}{2\pi}$. The magnetic field at the cente

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $L$ be the total length of the wire. For the circular coil of radius $R$,$L = 2\pi R$,so $R = \frac{L}{2\pi}$. The magnetic field at the center is $B_{C_1} = \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.For the square coil of side $a$,$L = 4a$,so $a = \frac{L}{4}$. The magnetic field at the center due to one side of length $a$ is $B_{side} = \frac{\mu_0 i}{4\pi(a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 i}{2\pi a} (2 \cdot \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 i}{\pi a}$.Since there are $4$ sides,the total magnetic field at the center is $B_{C_2} = 4 \cdot B_{side} = \frac{4\sqrt{2} \mu_0 i}{\pi a} = \frac{4\sqrt{2} \mu_0 i}{\pi (L/4)} = \frac{16\sqrt{2} \mu_0 i}{\pi L}$.The ratio is $\frac{B_{C_1}}{B_{C_2}} = \frac{\mu_0 i \pi / L}{16\sqrt{2} \mu_0 i / \pi L} = \frac{\pi^2}{16\sqrt{2}}$.Wait,re-evaluating the provided solution steps: $B_{C_2} = 4 \cdot \frac{\mu_0 i}{4\pi(a/2)} [\sin 45^{\circ} + \sin 45^{\circ}] = \frac{\mu_0 i}{\pi a/2} [\sqrt{2}] = \frac{2\sqrt{2} \mu_0 i}{\pi a}$. Substituting $a = \frac{\pi R}{2}$,we get $B_{C_2} = \frac{2\sqrt{2} \mu_0 i}{\pi (\pi R / 2)} = \frac{4\sqrt{2} \mu_0 i}{\pi^2 R}$.Then $\frac{B_{C_1}}{B_{C_2}} = \frac{\mu_0 i / 2R}{4\sqrt{2} \mu_0 i / \pi^2 R} = \frac{\pi^2}{8\sqrt{2}}$.

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

With a standard rectangular bar magnet,the time period of a vibration magnetometer is $4 ~s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

The three lines given by the combined equation $y^3 - 4x^2y = 0$ represent:

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then what is the set of possible values of $a$?

$A$ photon and an electron have equal energy $E$. The ratio $\lambda_{\text{photon}} / \lambda_{\text{electron}}$ is proportional to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module