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Step-by-Step Solution
Step 1: Identify the Relevant Concepts
When an electron is accelerated through a potential difference $V$, it gains kinetic energy equal to the electric potential energy lost. Once it enters a uniform magnetic field $B$ with velocity $v$ perpendicular to the field, it follows a circular path because of the Lorentz force $F = qvB$. The radius $r$ of this circular trajectory is given by the balance of centripetal and magnetic forces:
$ m v^2 / r = q v B \quad \Rightarrow \quad r = \frac{m v}{q B}.$
Step 2: Relate Electron's Velocity to the Accelerating Potential
The kinetic energy gained by the electron after being accelerated by a potential $V$ is
$ \frac{1}{2} m v^2 = eV,
where $e$ is the charge of the electron. Solving for $v$ gives
$ v = \sqrt{\frac{2 e V}{m}}.
Step 3: Substitute the Electron’s Velocity into the Radius Formula
Using $ v = \sqrt{\frac{2 e V}{m}} $ in the expression $ r = \frac{m v}{q B} $ (noting $q = e$ for an electron), we get:
$ r = \frac{m}{e B} \sqrt{\frac{2 e V}{m}}
= \frac{\sqrt{2 m e V}}{e B}.
Step 4: Plug in the Known Values
Electron mass, $ m = 9.1 \times 10^{-31}\,\text{kg}$
Electron charge, $ e = 1.6 \times 10^{-19}\,\text{C}$
Accelerating voltage, $ V = 500\,\text{V}$
Magnetic field, $ B = 100\,\text{mT} = 100 \times 10^{-3}\,\text{T} = 0.1\,\text{T}$
Substitute these into the expression for $r$:
$ r = \frac{\sqrt{2 \cdot 9.1 \times 10^{-31} \cdot \bigl(1.6 \times 10^{-19}\bigr) \cdot 500}}{\bigl(1.6 \times 10^{-19}\bigr)\, \cdot 0.1}.
Step 5: Evaluate the Expression
First, compute inside the square root:
$ \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 500}.
This simplifies numerically, and then dividing by $ (1.6 \times 10^{-19}) \times 0.1 $ gives:
$ r \approx 7.5 \times 10^{-4}\,\text{m}.
Final Answer
The radius of the path described by the electron is:
$\displaystyle 7.5 \times 10^{-4}\,\text{m}.$
