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Step-by-Step Solution
Step 1: State the Relevant Physical Situation
An electron moves in a circular path of radius $r$ with a uniform speed $v$, producing a magnetic field $B$ at the center of its circular orbit. We want to find how $r$ depends on $v$ and $B$.
Step 2: Write the Expression for the Magnetic Field
The magnetic field produced by a moving charge (considered as a current in a circular loop) at the center of the loop is given by
$ B = \frac{\mu_0 \, i}{2r}, $
where $r$ is the radius of the circular path, and $i$ is the effective current due to the moving electron.
Step 3: Determine the Effective Current
If the electron carries charge $q$ (magnitude $q = e$) and completes one revolution in time $T$, the current can be written as:
$ i = \frac{\text{charge per revolution}}{\text{time for one revolution}} = \frac{q}{T}. $
Since the electron moves with speed $v$ in a circular path of circumference $2\pi r$, the time for one revolution is
$ T = \frac{2\pi r}{v}. $
Thus,
$ i = \frac{q}{2\pi r / v} = \frac{q \, v}{2\pi r}. $
Step 4: Substitute the Expression for Current into the Magnetic Field Formula
Plugging $ i = \frac{q \, v}{2\pi r} $ into $ B = \frac{\mu_0 \, i}{2r} $, we get:
\[
B = \frac{\mu_0}{2r} \left(\frac{q \, v}{2\pi r}\right) = \frac{\mu_0 \, q \, v}{4\pi \, r^2}.
\]
Step 5: Solve for the Radius $r$
Rearrange the above equation to express $r$ in terms of $v$ and $B$:
\[
B = \frac{\mu_0 \, q \, v}{4\pi \, r^2} \quad \Longrightarrow \quad r^2 = \frac{\mu_0 \, q \, v}{4\pi \, B}.
\]
Therefore,
\[
r \propto \sqrt{\frac{v}{B}}.
\]
Step 6: Conclude the Proportionality
The direct conclusion is that
\[
r \propto \sqrt{\frac{v}{B}}.
\]
This corresponds to the correct choice: $ \sqrt{\frac{v}{B}}. $
