Two identical wires A and B,each of length l, | vedclass

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

Question

(B) For wire $A$ (circle): The circumference is $l = 2\pi R$,so $R = \frac{l}{2\pi}$. The magnetic field at the center is $B_A = \frac{\mu_0 I}{2R} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For wire $A$ (circle): The circumference is $l = 2\pi R$,so $R = \frac{l}{2\pi}$. The magnetic field at the center is $B_A = \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2(l/2\pi)} = \frac{\mu_0 I \pi}{l}$.For wire $B$ (square): The perimeter is $l = 4a$,so $a = \frac{l}{4}$. The distance from the center to a side is $d = \frac{a}{2} = \frac{l}{8}$. The magnetic field at the center due to one side is $B_1 = \frac{\mu_0 I}{4\pi d} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{4\pi (l/8)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{2\mu_0 I}{\pi l} \sqrt{2}$.Since there are $4$ sides,the total field is $B_B = 4 	imes B_1 = \frac{8\sqrt{2}\mu_0 I}{\pi l}$.Calculating the ratio: $\frac{B_A}{B_B} = \frac{\mu_0 I \pi / l}{8\sqrt{2}\mu_0 I / \pi l} = \frac{\pi^2}{8\sqrt{2}}$.

Similar Questions

$A$ current $i$ flows in a circular arc of wire of radius $R$,which subtends an angle of $3\pi / 2$ radians at its centre. The magnetic induction at the centre is

According to the Biot-Savart law,what is the magnetic field at a point on the axis of a current-carrying wire?

The magnetic field at a distance of $10 \ cm$ from a long straight thin wire carrying a current of $4 \ A$ is (in $\mu T$)

$A$ $10 \ A$ current is passing through a very long wire of radius $5 \ cm$. The magnetic field at a distance of $2 \ cm$ inside from its curved surface is . . . . . . $\times 10^{-5} \ T$.

$A$ coil having $9$ turns carrying a current produces a magnetic field $B_1$ at the centre. Now,the coil is rewound into $3$ turns carrying the same current. Then,the magnetic field at the centre $B_2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module