radius of path?

Question

an electron is moving at 2.5 * 10^6 m/s perpendicular to a 3.5 mT magnetic field. What is the radius of its path?

a) 5.1 cm
b) 3.9 cm
c) 1.6 cm
d) 4.1 cm
e) 2.0 cm

please help and explain thanks

Answer 1

Sounds like a cyclotron problem to me.

V/R= Bq/m
R= Vm/(Bq)
m=9.109 E-31 kg
q= -1.60 E–19 C
R= 2.5 E+6 x 9.109 E-31/(3.5 E-3 x 1.60 E–19)
R=0.41cm

Answer 2

First calculate your section for the 1st circle. If it makes issues easier, you could convert it into centimeters from the getgo. this would make our a million.4 a a hundred and forty. So: A1 = ?r^2 A1 = ?(a hundred and forty)^2 With this you will have your first section that we will call A1. Now calculate for the 2nd section by ability of including interior the extra 70 centimeters. A2 = ?(a hundred and forty + 70)^2 Now discover the adaptation between the two. A1 - A2 = A(walkway) in centimeters