Question

An insulated wire is wound so that it forms a flat coil with $N$ turns. The radii of the inner and outermost turns are $r_1$ and $r_2$ respectively. What will be the magnetic moment of this coil when a current $I$ flows through it ?

• A. $\frac{\pi NI}{3}(r_1^2+r_1{r}_2+r_2^2)$
• B. $\frac{\pi NI}{3}(r_1^2-r_1{r}_2+r_2^2)$
• C. $\frac{\pi NI}{3}(r_1^2+r_2^2)$
• D. $\frac{\pi NI}{3}(r_1^2-r_2^2)$

Answer 1

The correct option is A $\frac{\pi NI}{3}(r_1^2+r_1{r}_2+r_2^2)$


Considering an elementary turn of the coil of radius $r$ and thickness $dr$.

The magnetic moment of this turn is,

$dM=I\pi r^2$

So, the net dipole moment of the coil will be

$M=\int I\pi r^{2}dN.........\left(1\right)$

Here, $dN$ is the number of turns lying between radius $r$ and $dr$

Now, for the entire coil

$\frac{dN}{dr}=\frac{N}{r_2-r_1}$

$\Rightarrow dN=\frac{N}{r_2-r_1}dr$

So, equation $\left(1\right)$ becomes

$M={\int }_{r_1}^{r_2}\frac{\pi IN}{r_2-r_1}{r}^{2}dr=\frac{\pi IN}{r_2-r_1}[\frac{r_2^3-r_1^3}{3}]$

$\therefore M=\frac{\pi NI}{3}(r_1^2+r_1{r}_2+r_2^2)$

Hence, option $\left(A\right)$ is the correct answer.