A conducting ring of radius R and one turn is | vedclass

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

Question

(A) The magnetic dipole moment of a current-carrying loop is given by $m = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the are

Vedclass · Accepted Answer

$\frac{m}{2}$

Vedclass · Answer

(A) The magnetic dipole moment of a current-carrying loop is given by $m = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the loop.For the first case,$N_1 = 1$,$R_1 = R$,and $A_1 = \pi R^2$. Thus,$m = I \pi R^2$.When the wire of length $L$ is converted into a ring of $N_2 = 2$ turns,the circumference of the new ring is $2 	imes (2 \pi R_2) = L$. Since $L = 2 \pi R$,we have $4 \pi R_2 = 2 \pi R$,which gives $R_2 = \frac{R}{2}$.The new area is $A_2 = \pi R_2^2 = \pi (\frac{R}{2})^2 = \frac{\pi R^2}{4}$.The new magnetic dipole moment $m'$ is $m' = N_2 I A_2 = 2 	imes I 	imes \frac{\pi R^2}{4} = \frac{I \pi R^2}{2}$.Since $m = I \pi R^2$,we get $m' = \frac{m}{2}$.

Similar Questions

$A$ 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.

$A$ thin circular wire carrying current $I$ has magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have magnetic moment $M_1$. The ratio $M$ to $M_1$ is

$A$ circular loop having radius $r$,carrying current $I$,produces a magnetic field at the centre of the loop equal to $B$. What will be the magnetic dipole moment of this loop?

$2 \ A$ current is flowing through two different small circular copper coils having radii in the ratio $1:2$. The ratio of their respective magnetic moments will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module