Q: A mono energetic electron beam with a speed of 5.2 × 10^{-6} m/s enters into a magnetic field 3 × 10^{-4} T to the beam.

Find the radius of the circle normal to the beam. Find the radius of the circle traced by the beam

(take e/m = 1.76 × 10^{11} c/kg)

Sol: From the equation, B e v = mv^{2}/r

Radius of circle , $\large r = \frac{m v}{e B} = \frac{v}{(e/m) B}$

$\large r = \frac{m v}{e B} = \frac{5.2 \times 10^6}{(1.76 \times 10^{11}) \times 3 \times 10^{-4}}$

r = 0.1 m.