Question

What is the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic fields. E--> and B--> are also perpendicular to each other and have magnitudes 7900V/m and 9.1*10^-3T , respectively.
What is the radius of the electron orbit if the electric field is turned off?

Answer 1

Explanation:

Given that,

Electric field, E = 7900 V/m

Magnetic field, $B=9.1\times 10^{-3}\ T$

(a) Let v is the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic fields. The magnitude of electric force and the magnetic force balance each other as :

$qE=qvB$

$v=\dfrac{E}{B}$

$v=\dfrac{7900\ V/m}{9.1\times 10^{-3}\ T}$

$v=8.68\times 10^5\ m/s$

(b) Let r is the radius of the electron orbit. The radius of the motion of electron is given by :

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\ kg\times 8.68\times 10^5\ m/s}{1.6\times 10^{-19}\ C\times 9.1\times 10^{-3}\ T}$

$r=5.4\times 10^{-4}\ m$

Hence, this is the required solution.