Question

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic field of 4.62×10−3t with both fields normal to the beam and to each other, produces no deflection of the electrons?When the electric field is removed, what is the radius of the electron orbit?What is the period of the orbit?

Answer 1

1) $3.38\cdot 10^6 m/s$

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

$F_E = F_B\\qE = qvB$

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

$v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s$

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

$qvB=m\frac{v^2}{r}$

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

$r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm$

3) $7.6\cdot 10^{-9}s$

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, $2 \pi r$, and the period, T:

$v=\frac{2\pi r}{T}$

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

$T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s$