Question

An electron of energy 10 eV is revolving round a circular path in a uniform magnetic field of \( 10^{-5} \) tesla. Determine the radius of the circular path.
(Given: mass of electron \( m = 9.1\times10^{-31} \) kg, charge \( e = 1.6\times10^{-19} \) C)

Hint:

The magnetic force provides the centripetal force, so \( r = \dfrac{mv}{qB} \). Get \( v \) from \( E = \tfrac{1}{2}mv^2 \), or use \( r = \dfrac{\sqrt{2mE}}{qB} \) directly.

Answer 1

Step 1 (Concept): When a charged particle enters a uniform magnetic field perpendicular to it, the magnetic force \( qvB \) supplies the centripetal force needed for circular motion. So
\[ qvB = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{qB} \]
Step 2 (Convert the energy to joule): The kinetic energy is
\[ E = 10\ \text{eV} = 10\times1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
Step 3 (Find the speed): Kinetic energy \( E = \tfrac{1}{2}mv^2 \), so
\[ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2\times1.6\times10^{-18}}{9.1\times10^{-31}}} \]
\[ v = \sqrt{3.516\times10^{12}} = 1.875\times10^{6}\ \text{m/s} \]
Step 4 (Substitute in the radius formula):
\[ r = \frac{mv}{qB} = \frac{(9.1\times10^{-31})(1.875\times10^{6})}{(1.6\times10^{-19})(10^{-5})} \]
Step 5 (Arithmetic): Numerator \( = 1.706\times10^{-24} \); denominator \( = 1.6\times10^{-24} \).
\[ r = \frac{1.706\times10^{-24}}{1.6\times10^{-24}} = 1.07\ \text{m} \]
\[\boxed{r \approx 1.07\ \text{m}}\]