Step 1: Understanding the Concept:
A moving charge behaves like an electric current.
When it rotates in a circle, it acts as a circular current loop, creating a magnetic field at its center.
Step 2: Key Formula or Approach:
Equivalent current $I = q \cdot f$, where $q$ is charge and $f$ is frequency.
Magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2r}$.
We are given $B = x \cdot \mu_0$. We equate the two to solve for $r$.
Step 3: Detailed Explanation:
Given charge $q = 1000e = 1000 \times 1.6 \times 10^{-19} \text{ C} = 1.6 \times 10^{-16} \text{ C}$.
Frequency of rotation $f = 1 \text{ rotation/second} = 1 \text{ Hz}$.
Calculate equivalent current:
\[ I = qf = (1.6 \times 10^{-16} \text{ C})(1 \text{ s}^{-1}) = 1.6 \times 10^{-16} \text{ A} \]
The magnetic field at the center is given as $B = x\mu_0$, where $x = 2 \times 10^{-16}$.
\[ B = 2 \times 10^{-16} \cdot \mu_0 \]
Using the formula for magnetic field of a loop:
\[ \frac{\mu_0 I}{2r} = 2 \times 10^{-16} \cdot \mu_0 \]
Cancel $\mu_0$ from both sides:
\[ \frac{I}{2r} = 2 \times 10^{-16} \]
Substitute the value of $I$:
\[ \frac{1.6 \times 10^{-16}}{2r} = 2 \times 10^{-16} \]
Cancel the $10^{-16}$ terms:
\[ \frac{1.6}{2r} = 2 \]
\[ \frac{0.8}{r} = 2 \]
\[ r = \frac{0.8}{2} = 0.4 \text{ m} \]
Step 4: Final Answer:
The radius '$r$' is 0.4 m.