Question

An electron with a speed of 3.55

Answer 1

a)The charge experiances a magnetic force only thru "vy". This force "qvyB" is center seeking and produces circular motion with centripetal force mvy^2/r as;

qvyB = mvy^2/r

r = mvy/qB (you can do the numbers)

With speed "vy" , the charge will travel thru one circumference "2pir" in a time "t" as;

vy = 2pir/t

t = 2pir/vy

using the result from above part this reduces to;

t = 2pim/qB

In this time "t" the unchanged horizontal velocity "vx" will move the charge a distance "P" as;

P = vxt = 2pimvx/qB