Answer to Question #89921 in Electricity and Magnetism for Shivam Nishad

Question

A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields of magnitudes 6 kVm^-1 and 10^-3 T, respectively. By maintaining the same magnetic field when the electric field is cut off, the electrons move in the magnetic field in a circular path. Calculate the speed of electrons and the radius of the circular path.

Answer 1

The absence of deflection in the first part of the problem means that both electric and magnetic forces act along the same line but in the opposite directions, i.e.

$$eE=evB \, \Rightarrow \, v = \frac{E}{B}$$

Substituting the numerical values, we obtain

$$v = \frac{6 \cdot 10^3}{10^{-3}} = 6 \cdot 10^6 \, m/s$$

In case when only the magnetic field is maintained, the condition of the circular orbit states:

$$evB = \frac{m v^2}{R} \, \Rightarrow \, R = \frac{m v}{eB} = \frac{m E}{e B^2}$$

Substituting the numerical values, we obtain

$$R = \frac{9.1 \cdot 10^{-31} \cdot 6 \cdot 10^{3}}{1.6 \cdot 10^{-19} \cdot 10^{-6}} \approx 3.4 \cdot 10^{-2} \, m = 3.4 \, cm$$

Answer: 6x10⁶ m/s; 3.4 cm