Electron Orbital Radius in Magnetic Field

Electron Orbital Radius in Magnetic Field (1996)

Question: A 10 eV electron circulates perpendicular to a magnetic field \( B = 10^{-4} \, \text{T} \). The orbital radius is:

(a) 11 cm

(b) 18 cm

(c) 12 cm

(d) 16 cm

Solution

Step 1: Convert energy to velocity
Kinetic energy \( E = 10 \, \text{eV} = 10 \times 1.602 \times 10^{-19} \, \text{J} = 1.602 \times 10^{-18} \, \text{J} \).
Using \( E = \frac{1}{2}mv^2 \): \[ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 1.602 \times 10^{-18}}{9.109 \times 10^{-31}}} = 1.876 \times 10^6 \, \text{m/s} \]

Step 2: Calculate radius
\[ r = \frac{mv}{qB} = \frac{(9.109 \times 10^{-31})(1.876 \times 10^6)}{(1.602 \times 10^{-19})(10^{-4})} = \frac{1.709 \times 10^{-24}}{1.602 \times 10^{-23}} = 0.1067 \, \text{m} = 10.67 \, \text{cm} \] Rounded to 11 cm.

Answer: Option (a) is correct. The magnetic field causes circular motion with radius ~11 cm.