Question

The ratio of magnetic field at the centre of a current carrying coil to its magnetic moment is $x$. If the current and radius both are doubled, the new ratio will become:

• A. $2x$
• B. $4x$
• C. $x/4$
• D. $x/8$

Answer 1

The correct option is D $x/8$

Magnetic field at center of current carrying coil is,
$B=\frac{{\mu }_{o}i}{2r}$

Magnetic moment is given by:

$M=i\pi r^2$
Given ratio is:

$x=\frac{B}{M}$

$x=\frac{{\mu }_o}{2\pi r^3}$
On doubling the current and radius,
$x^{\prime }=\frac{{\mu }_0}{2\pi \times (2r{)}^3}$
$\therefore x^{\prime }=\frac{x}{8}$