A particle carrying a charge equal to 1000 times the charge on an electron, is rotating one rotation per second in a circular path of radius ' r ' m . If the magnetic field produced at the centre of the path is $x$ times the permeability of vacuum, the radius ' $r$ ' in $m$ is $\left[\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right] \quad\left[\mathrm{x}=2 \times 10^{-16}\right]$

Finding the radius of a charged particle's circular path

The magnetic field at the center of a circular loop carrying current $I$ with radius $r$ is given by $B = \frac{\mu_0 I}{2r}$.

A charge $Q = 1000e = 1.6 \times 10^{-16}$ C rotating with frequency $f = 1$ Hz produces an effective current $I = Qf = 1.6 \times 10^{-16}$ A.

Given $B = x\mu_0$ where $x = 2 \times 10^{-16}$, substitution yields $x\mu_0 = \frac{\mu_0 I}{2r}$.

Canceling $\mu_0$ and solving for $r$ gives $r = \frac{I}{2x} = \frac{1.6 \times 10^{-16}}{2 \times (2 \times 10^{-16})} = \frac{1.6}{4} = 0.4$ m.

The radius is therefore $0.4$ m, corresponding to answer choice D.