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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
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Verified Answer
The correct answer is:
$x / 8$
$$
\begin{aligned}
& \text { Hints: } \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 a} \quad \quad \mathrm{M}=\mathrm{I}\left(\pi a^2\right) \\
& \frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_0 \mathrm{I}}{2 a} \times \frac{1}{\mathrm{I} \pi a^2}=\frac{\mu_0}{2 \pi a^3}=x
\end{aligned}
$$
Again, Ratio $=\frac{\mu_0}{2 \pi(2 a)^3}=\frac{1}{8}\left(\frac{\mu_0}{2 \pi a^3}\right)=\frac{x}{8}$
\begin{aligned}
& \text { Hints: } \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 a} \quad \quad \mathrm{M}=\mathrm{I}\left(\pi a^2\right) \\
& \frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_0 \mathrm{I}}{2 a} \times \frac{1}{\mathrm{I} \pi a^2}=\frac{\mu_0}{2 \pi a^3}=x
\end{aligned}
$$
Again, Ratio $=\frac{\mu_0}{2 \pi(2 a)^3}=\frac{1}{8}\left(\frac{\mu_0}{2 \pi a^3}\right)=\frac{x}{8}$
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