Period of an Electron Around a Magnetic Field

AI Thread Summary
The discussion focuses on calculating the period of an electron's motion in a uniform magnetic field of 0.152 teslas. The initial attempt incorrectly used the speed of light for the electron's velocity, leading to an erroneous period of 1.86E-6 seconds. The correct approach involves manipulating the equations to eliminate the radius and derive the period directly from the magnetic force and mass of the electron. By substituting the appropriate values, the correct period is found to be 2.35E-10 seconds. This solution emphasizes the importance of using accurate velocities for electrons in magnetic fields.
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Homework Statement



An electron enters a uniform magnetic field of 0.152 teslas such that the electron follows a circular path. What is the period of the electron's motion? Electrons have a mass of about 9.11 x 10-31 kilograms each. Hint: Since the radius of the electron's path is not given, it must cancel out of the equations.


Homework Equations



Fm = Q * V * B

F = M * A

A = V^2/r

V= 2(pi)r/T


The Attempt at a Solution



Q * V * B = M * A
Q * V * B = M * V^2/r
Q * 2(pi)r/T * B = M* V^2/r
Q * 2(pi)/T * B = M * V^2
(1.6E-19) * 2(pi)/T = (9.11E-31) * (3.00E8)^2
Solve for T
T = 1.86E-6 s
But it is wrong...
 
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How can you use v=3.00E8? Electrons can't possibly travel at the speed of light, and even if they could, the question doesn't say that they do.

Try substituting 2(pi)r/T in the v on the right side as well. Better yet, cancel out one v in Q * V * B = M * V^2/r, then substitute 2pi*r/T into the remaining v.
 
Cool got it... here is what I did if somebody else would like to see...

Q * V * B = m * v^2/r

Q * B = M * V / R

Q* B = M * (2(pi)r/T)/r

Q * B = M * (2pi)/T

T = 2(pi) * M / Q * B

I got 2.35E-10 seconds... worked great... thanks
 
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