Question

A)What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic fields of magnitude 6800 v/m and 7.8×10-3T , respectively?
B)What is the radius of the electron orbit if the electric field is turned off?

Answer 1

E = vB
v = E / B = 6800 / 7.8e-3
v =8.7e5 m /s

B.
mv^2 / r = qvB
mv / r = qB
r = mv / [ qB]
r = 9.1e-31 * 8.7e5 / [1.6e-19 * 7.8e-3]
r= 6.35 e-4 m

Answer 2

A)
qE = q(v x B)
E=vBsin90
v = E/B = 6800 / 7.8*10^-3
v =8.7*10^5 m /s

B.
centripetal force=magnetic force
mv^2 / r = qvBsin90
mv / r = qB
r = mv / qB
r = 9.1*10^-31 * 8.7*10^5 /(1.6*10^-19 * 7.8*10^-3)
r= 6.35 *10^-4 m