A circular loop and a square loop are formed | 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 the circular loop,the circumference is $2 \pi r = l$,so the radius $r = \frac{l}{2 \pi}$.The magnetic dipole

Vedclass · Accepted Answer

$\frac{4}{\pi}$

Vedclass · Answer

(B) Let the length of the wire be $l$.For the circular loop,the circumference is $2 \pi r = l$,so the radius $r = \frac{l}{2 \pi}$.The magnetic dipole moment of the circular loop is $M_{1} = i A_{1} = i \pi r^{2}$.Substituting $r$,we get $M_{1} = i \pi \left(\frac{l}{2 \pi}\right)^{2} = \frac{i l^{2}}{4 \pi}$.For the square loop,the perimeter is $4 a = l$,so the side length $a = \frac{l}{4}$.The magnetic dipole moment of the square loop is $M_{2} = i A_{2} = i a^{2}$.Substituting $a$,we get $M_{2} = i \left(\frac{l}{4}\right)^{2} = \frac{i l^{2}}{16}$.The ratio of the dipole moments is $\frac{M_{1}}{M_{2}} = \frac{i l^{2} / 4 \pi}{i l^{2} / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

$A$ circular loop and a square loop are formed from the same wire and the same current is passed through them. Find the ratio of their dipole moments.

Similar Questions

Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of current and the thumb gives the direction of the magnetic field. In this case,the upper side of the loop may be thought of as

What is the magnetic dipole moment for a coil? Write its $SI$ unit and dimensional formula,and explain its stable and unstable equilibrium.

The magnetic moment is $NOT$ associated with

Derive an expression for the torque acting on a current-carrying loop suspended in a uniform magnetic field.

Magnetic induction at the centre of a circular loop of area $\pi \ m^2$ is $0.1 \ T$. The magnetic moment of the loop is ( $\mu_0 = \text{permeability of air}$ ).

Vedclass Test Series

Exam Paper Generator

Online Exam Module