Step 1: Establish the force balance for circular motion.
When a charged particle moves in a magnetic field, the magnetic force provides the centripetal force: \[ qvB = \frac{mv^2}{r} \implies B = \frac{mv}{qr} = \frac{p}{qr} \] where $p = mv$ is the linear momentum of the particle.
Step 2: Relate momentum to kinetic energy.
For a non-relativistic particle: $K = \dfrac{p^2}{2m}$, so $p = \sqrt{2mK}$. Multiplying both sides by $c$: \[ pc = \sqrt{2mc^2 K} \]
Step 3: Substitute the given values in energy units.
Given: $mc^2 = 0.5\,\text{MeV} = 5 \times 10^5\,\text{eV}$ and $K = 100\,\text{eV}$: \[ pc = \sqrt{2 \times 5 \times 10^5 \times 100} = \sqrt{10^8}\,\text{eV} = 10^4\,\text{eV} \] So $p = \dfrac{10^4\,\text{eV}}{c}$.
Step 4: Convert momentum to SI units.
\[ p = \frac{10^4 \times 1.6 \times 10^{-19}}{3 \times 10^8}\,\text{kg m/s} = \frac{1.6 \times 10^{-15}}{3 \times 10^8} = 5.33 \times 10^{-24}\,\text{kg m/s} \]
Step 5: Calculate the magnetic field.
With $q = 1.6 \times 10^{-19}\,\text{C}$ and $r = 10\,\text{cm} = 0.1\,\text{m}$: \[ B = \frac{p}{qr} = \frac{5.33 \times 10^{-24}}{1.6 \times 10^{-19} \times 0.1} = \frac{5.33 \times 10^{-24}}{1.6 \times 10^{-20}} \approx 3.3 \times 10^{-4}\,\text{T} \]
Step 6: State the final answer.
The magnitude of the magnetic field is: \[ \boxed{|\vec{B}| \approx 3.3 \times 10^{-4}\,\text{T}} \]