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

qE= qVB

E= vB

7900= v * 9.1*10^-3

v= 868131 m/s

r= mv/qB= 9.1*10^-31* 868131/ 1.6*10 ^-19 * 9.1* 10^-3 = 5.426 * 10 ^-4 m= 0.5426 mm

Answer 2

If the beam of electrons goes unde?ected, the force on the electron due to the electric force must be the
same magnitude as the force on the electron due to the magnetic force. Therefore, qE = qvB, and solving
for v we ?nd that
v =E/B
v = 7900 V/m/(9.1* 10^?3) T
= 86831 m/s.
If the electric ?eld is turned o?, the magnetic force will cause the electron to undergo uniform circular
motion, so qvB = mv2/r. Solving for r we ?nd that
r =mv/qB

=> r = (9.11 * 10^(-31) *86831 )/(1.6*10^(-19) * 9.1*10^(-3)) = 5.4 * 10^(-5) m