posted by J on .
An electron beam of a TV picture tube is accelerated through a potential difference of 2.00 kV. It then passes into a magnetic field perpendicular to its path, where it moves in a circular diameter of .360 m. What is the magnitude of the field?
I am lost. I keep getting the wrong answer! Help!
First calculate the velocity V after being accelerated by the potential difference.
(1/2)m V^2 = 2000 J/coulomb*1.602*10^-19 coulombs
Here, m is the electron mass
When travelling arouind the B field line,
e V B = m V^2/R , the centripetal force that keeps it moving in a circle.
B = (m/e)(V/R)