Question

$B$ is the magnetic field at the centre of a circular loop. The magnetic moment of the circular loop of area $A$ is found to be $N\frac{B{A}^Q}{{\mu }_0\sqrt{\pi }}$ . The value of $2\left(N+Q\right)$ is -------.

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 $Q=1.5$
$\therefore 2\left(N+Q\right)=7$