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

Answer: 7

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
$B=\frac{{\mu }_{0}I}{2r}$ $\Rightarrow B$
$=\frac{{\mu }_{0}I}{2\sqrt{\frac{A}{\pi }}}\Rightarrow I=\frac{2B}{{\mu }_0}\sqrt{\frac{A}{\pi }}$
Now, magnetic moment of the loop is
$M=IA$
$=\frac{2B}{{\mu }_0}\sqrt{\frac{A}{\pi }}\cdot A$
$=\frac{2B}{{\mu }_0}\frac{A^{\frac{3}{2}}}{\sqrt{\pi }}$
$\Rightarrow n=2$ and $m=1.5$
$\therefore n+m=3.5$