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

Answer: (A)

$\frac{x }{ 8}$

Answer 2

The magnetic field at the centre of a current carrying loop is given by
$B=\frac{\mu_{0}}{4 \pi}\left(\frac{2 \pi i}{a}\right)=\frac{\mu_{0} i}{2 a}$
The magnetic moment at the centre of current carrying loop is given by $M=i\left(\pi a^{2}\right)$
Thus, $\frac{B}{M}=\frac{\mu_{0} \dot{i}}{2 a} \times \frac{1}{i \pi a^{2}}=\frac{\mu_{0}}{2 \pi a^{3}}=x$ (given)
When both the current and the radius are doubled, the ratio becomes
$\frac{\mu_{0}}{2 \pi(2 a)^{3}}=\frac{\mu_{0}}{8\left(2 \pi a^{3}\right)}=\frac{x}{8}$
So, the correct option is (A) : $\frac{x}{8}$.