(1) (2 points) Electron A is fired horizontally with speed 1 Mm/s into a region
ID: 2034366 • Letter: #
Question
(1) (2 points) Electron A is fired horizontally with speed 1 Mm/s into a region where a vertical magnetic field exists. Electron B is fired along the same path with speed 2 Mm/s. (a) Which electron has a larger magnetic force exerted upon it? Or, are the forces equal? Or, are the forces zero? b) Which electron follows a path that curves more sharply? Or, do both electrons follow the same curved path? Or, do both electrons follow a straight line path? Show how method of ratios can be used to determine the answer to each question.Explanation / Answer
Given,
v = 1 x 10^6 m/s ; v' = 2 x 10^6 m/s
a)We know that the magnetic force acting will be:
Fm = q v B sin(theta)
We see from the expression that the magnetic force is directly proportional to velocity
Fm-A = q v B sin(theta)
Fm-B = q v' B sin(theta)
Fm-B/Fm-A = v'/v = 2 x 10^6 m/s/(1 x 10^6) = 2
Fm-B = 2 Fm-A
So, electron B will have larger magnitude of magnetic force on it than electron A.
b)the centripital force balances the magnetic force
q v B = m v^2/R
R = m v/ B q
R(A) = m v/B q
R(B) = m v'/B q = m (2v)/Bq
R(A)/R(B) = v/v' = 1/2
Smaller the radius more sharp is the curve.
A will have smaller radius than B and hence will follow a more sharp curve.
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