Question

The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is x. When both the current and radius is doubled the ratio will be

• A. $\frac{x}{8}$
• B. $\frac{x}{4}$
• C. $\frac{x}{2}$
• D. $2x$

Answer 1

The correct option is A $\frac{x}{8}$

The magnetic field at the centre of a current carrying loop is given by
$B=\frac{{\mu }_0}{4\pi }\left(\begin{array}{c}\frac{2\pi i}{a}\end{array}\right)=\frac{{\mu }_{0}i}{2a}$
The magnetic moment at the centre of current carrying loop is given by $M=i\left(\pi a^2\right)$
Thus, $\frac{B}{M}=\frac{{\mu }_{o}i}{2a}\times \frac{1}{i\pi a^2}=\frac{{\mu }_0}{2\pi a^3}=x$ (given)
When both the current and the radius are doubled, the ratio becomes
$\frac{{\mu }_0}{2\pi (2a{)}^3}=\frac{{\mu }_0}{8\left(2\pi a^3\right)}=\frac{x}{8}$