Question

An electron with kinetic energy $ 100\, \text{eV} $ moves in a circular path of radius $ 10\, \text{cm} $ in a magnetic field. Find the approximate magnitude of the magnetic field $ B $.
[Given: mass of electron = $ 0.5\, \text{MeV}/c^2 $]

• A. \( 3.3 \times 10^{-4}\, \text{T} \)
• B. \( 2.6 \times 10^{-4}\, \text{T} \)
• C. \( 1.7 \times 10^{-4}\, \text{T} \)
• D. \( 4.3 \times 10^{-4}\, \text{T} \)

Hint:

Use \( B = \frac{mv}{qR} \), and find \( v \) from kinetic energy: \( v = \sqrt{2K.E/m} \)

Answer 1

The Correct Option is A

Use the formula: \[ B = \frac{mv}{qR} \] From kinetic energy: \[ K.E = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2K.E}{m}} \] Given: \[ K.E = 100\, \text{eV} = 100 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-17}\, \text{J} \] \[ m = \frac{0.5\, \text{MeV}}{c^2} = \frac{0.5 \times 10^6 \times 1.6 \times 10^{-19}}{(3 \times 10^8)^2} \approx 9.1 \times 10^{-31}\, \text{kg} \] So, \[ v = \sqrt{\frac{2 \cdot 1.6 \times 10^{-17}}{9.1 \times 10^{-31}}} \approx 1.88 \times 10^6\, \text{m/s} \] \[ B = \frac{9.1 \times 10^{-31} \cdot 1.88 \times 10^6}{1.6 \times 10^{-19} \cdot 0.1} \approx 3.3 \times 10^{-4}\, \text{T} \]