Q134 / 160 · WBJEE 2010 · Physics

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

  • Correct answer
    A
    $x / 8$
  • B
    $x / 4$
  • C
    $x / 2$
  • D
    $2 x$

Solution

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

Hence, the correct answer is option: 'A'.

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