Question

If $B$ is the magnetic field at the centre of a circular loop of area $A$ and its magnetic dipole moment is found to be $n\frac{B A^{m}}{\mu _{0} \sqrt{\pi }}$ , then what is the value of $2\times \left(n + m\right)$ ?

Answer 1

Let r be the radius of the circular loop.
$\therefore A=\pi r^{2}$
$\Rightarrow r=\sqrt{\frac{A}{\pi }}$
Magnetic field at the centre of the loop is
$B=\frac{\mu _{0} I}{2 r}$ $\Rightarrow B$
$=\frac{\mu _{0} I}{2 \sqrt{\frac{A}{\pi }}}\Rightarrow I=\frac{2 B}{\mu _{0}}\sqrt{\frac{A}{\pi }}$
Now, magnetic moment of the loop is
$M=IA$
$=\frac{2 B}{\mu _{0}}\sqrt{\frac{A}{\pi }}\cdot A$
$=\frac{2 B}{\mu _{0}}\frac{A^{\frac{3}{2}}}{\sqrt{\pi }}$
$\Rightarrow n=2$ and $m=1.5$
$\therefore n+m=3.5$