The magnetic field at the centre of a current | vedclass

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

Question

(A) Let $r$ be the radius of the circular loop. Since the area $A = \pi r^2$,we have $r = \sqrt{\frac{A}{\pi}}$. The magnetic field at the centre of a

Vedclass · Accepted Answer

$\frac{2 B A^{3 / 2}}{\mu_0 \pi^{1 / 2}}$

Vedclass · Answer

(A) Let $r$ be the radius of the circular loop. Since the area $A = \pi r^2$,we have $r = \sqrt{\frac{A}{\pi}}$. The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2r}$. Substituting the value of $r$,we get $B = \frac{\mu_0 I}{2 \sqrt{A/\pi}}$. Solving for current $I$,we get $I = \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}}$. The magnetic moment $M$ of the loop is defined as $M = I A$. Substituting the expression for $I$,we get $M = \left( \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}} \right) A = \frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$. Thus,the correct option is $A$.

The magnetic field at the centre of a current carrying circular coil of an area $A$ is $B$. The magnetic moment of the coil is (where $\mu_0$ is the permeability of free space).

Similar Questions

$A$ current of $3 \, A$ is flowing in a plane circular coil of radius $4 \, cm$ and number of turns $20$. The coil is placed in a uniform magnetic field of magnetic induction $0.5 \, T$. The magnetic dipole moment of the coil is $... \, A \cdot m^2$.

The number of turns of two circular coils $A$ and $B$ are $300$ and $200$ respectively. The magnetic moments of the two coils $A$ and $B$ are in the ratio $1:2$. If the two coils carry equal currents, then the ratio of radii of coils $A$ and $B$ is

The magnitude of magnetic induction at a point on the axis at a large distance $(r)$ from the centre of a circular coil of $n$ turns and area $A$ carrying current $I$ is given by:

The dipole moment of a circular loop carrying a current $I$ is $m$ and the magnetic field at the centre of the loop is $B_1$. When the dipole moment is doubled by keeping the current constant,the magnetic field at the centre of the loop is $B_2$. The ratio $\frac{B_1}{B_2}$ is

$A$ current loop placed in a magnetic field behaves like a

Vedclass Test Series

Exam Paper Generator

Online Exam Module