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 (Set up): For a charge \(q\) moving with speed \(v\) perpendicular to a field \(B\), the circular radius is \(r = \dfrac{mv}{qB}\). Instead of finding \(v\) first, we can express \(v\) through the energy in one shot.
Step 2 (Link speed to energy): Since \(E = \tfrac{1}{2}mv^2\), we get \(mv = \sqrt{2mE}\). Substituting this into the radius formula gives a single compact expression:
\[ r = \frac{\sqrt{2mE}}{qB} \]
Step 3 (Insert values): Here \(E = 10\ \text{eV} = 1.6\times10^{-18}\) J, \(m = 9.1\times10^{-31}\) kg, \(q = 1.6\times10^{-19}\) C, \(B = 10^{-5}\) T.
\[ 2mE = 2\times9.1\times10^{-31}\times1.6\times10^{-18} = 2.912\times10^{-48} \]
\[ \sqrt{2mE} = 1.706\times10^{-24}\ \text{kg·m/s} \]
Step 4 (Divide by \(qB\)): \(qB = 1.6\times10^{-19}\times10^{-5} = 1.6\times10^{-24}\).
\[ r = \frac{1.706\times10^{-24}}{1.6\times10^{-24}} = 1.07\ \text{m} \]
The radius comes out the same, confirming the result.
\[\boxed{r \approx 1.07\ \text{m}}\]