Question

What is the velocity of a beam of electrons that go undeflected when passing through crossed (perpendicular) electric and magnetic fields of magnitude 1.63×104V/m and 2.50×10−3T, respectively? me=9.11×10−31kg, e=1.60×10−19C.

Answer 1

Answer: V = 67808m/s

Step-by-step explanation:

Parameter given

E = 1.63*104V/m

B = 2.50*10-3T

V = E/B

Velocity of the beam of electron = (Electric field/Magnetic field)

V = (1.63*104V/m) ÷ (2.5*10-3T)

= 169.52/ 0.0025

V= 67808m/s