A proton moving freely in a circular path perpendicular to a constant magnetic field takes 0.80 µs....?

Question

Homework help for physics... Can you show me the steps? Thanks!

A proton moving freely in a circular path perpendicular to a constant magnetic field takes 0.80 µs to complete one revolution. Determine the magnitude of the magnetic field.

T

Answer 1

Step 1: The force on the proton is 'F'

F = q v B

where q = charge of the proton

v = velocity of the proton

B = magentic field

Step 2: From Newton's second law the force supplies an acceleration to keep the proton on a circular path

F = ma

m = mass of the proton

a = acceleration

Step 3: From uniform circular motion:

a = v^2/r

r = radius of the circle

Step 4: Combine the equations and solve for B:

F = q v B = m v^2 / r

B = m v / (r q)

Step 5: Express in terms of the period:

v / r = omega (the angular velocity)

and omega = 2 pi / T

where T is the period

B = 2 pi m / (q T)

Step 6: solve for B

m = 1.67 x 10^-27 kg

q = 1.60 X 10^-19 coul

B = 2 (pi) (1.67 x 10^-27) / (1.60 x 10^-19) / (0.8 x 10^-6)

B = 0.082 T