Q149 / 180 · KCET 2020 · Physics
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$, if the current and the radius both are doubled. The new ratio will become
- A$2 x$
- B$4 x$
- C$\frac{x}{4}$
- Correct answerD$\frac{x}{8}$
Solution
Magnetic field at the centre of current carrying circular coil
$B=\frac{\mu_{0} I}{2 r}$
where, $I=$ current and $r=$ radius.
Magnetic dipole moment,
$\begin{aligned}
& m=I A=I \cdot \pi r^{2} \\
\text { Given, } \quad & \frac{B}{m}=x \\
\Rightarrow \quad & \frac{\mu_{0} I}{2 r} \\
\Rightarrow \quad & \frac{\mu_{0}}{\pi I r^{2}}=x ...(i)\\
\end{aligned}$
When current and radius, both are doubled, then
$\begin{aligned} \frac{B^{\prime}}{m^{\prime}} &=\frac{\frac{\mu_{0} 2 I}{2 \cdot 2 r}}{\pi(2 I)(2 r)^{2}} \\ &=\frac{\frac{\mu_{0} I}{2 r}}{8 \pi r^{2} I}=\frac{\mu_{0}}{16 \pi r^{3}} \\ &=\frac{1}{8} \cdot \frac{\mu_{0}}{2 \pi r^{3}}=\frac{x}{8} \quad \text { [from Eq. (i)] } \end{aligned}$
Hence, the correct answer is option: 'D'.