$A$ wire of length $L$ metre carrying a current of $I$ ampere is bent in the form of a circle. Its magnetic moment is:

  • A
    $\frac{I L^{2}}{4 \pi} \; A \cdot m^{2}$
  • B
    $\frac{I L^{2}}{4} \; A \cdot m^{2}$
  • C
    $\frac{I \pi L^{2}}{4} \; A \cdot m^{2}$
  • D
    $\frac{2 I L^{2}}{\pi} \; A \cdot m^{2}$

Explore More

Similar Questions

$A$ current-carrying loop consists of $3$ identical quarter circles of radius $R$,lying in the positive quadrants of the $xy$,$yz$,and $zx$ planes with their centers at the origin,joined together. Find the direction and magnitude of the magnetic field $\vec{B}$ at the origin.

$A$ wire carrying current $I$ is bent in the shape $A, B, C, D, E, F, A$ as shown,where rectangle $A, B, C, D, A$ and $A, D, E, F, A$ are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b$,then the magnitude and direction of the magnetic moment of the loop $A, B, C, D, E, F, A$ is

$A$ conducting ring of radius $R$ and one turn is formed from a conducting wire of length $L$ and on passing current $I$ the obtained magnetic dipole moment is $m$. If this wire is then converted to a ring of two turns and on passing electric current $I$,the new magnetic dipole moment obtained is . . . . . . .

$A$ wire of length $2\,m$ is bent into a circular loop. When a current of $1\,A$ is passed through the loop,then the magnetic moment of the loop is

Difficult
View Solution

An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo