- #1
Dominic90
- 3
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- Homework Statement
- An electron, accelerated from rest by a potential difference ∆V, enters a magnetic field B, perpendicular to the lines of force. The electron therefore travels a circular trajectory of radius r. If ∆V is doubled and B halved, what will be the radius R, as a function of r, of the new trajectory?
The same electron is now in uniform circular motion of radius 26.1 𝜇𝑚 subject to a field uniform magnetic. The magnetic force acting on it has an intensity of 1.60 ∙ 10^(−17)𝑁.
Calculate the kinetic energy of the electron.
- Relevant Equations
- K = (1/2) mv²
E = q ∆V
r = m v / (|q|B)
F = |q|vB
Centripetal force = mv²/r
Hi, I tried to solve this exercise but I'm not sure about the process.
First of all, I imposed that "K = E":
so that "v = √ ( (2q ∆V)/m))"
then I replaced in "r = m v / (| q |B)", v with "√ ( (2q ∆V)/m))", and found out that R = (2√(2)) r.
Then for the second point,
I imposed Lorenz Force Law = Centripetal force and isolated v.
Then I substituted v in K
and obtained that "K= (1/2 ) F r".
Is this correct?
First of all, I imposed that "K = E":
so that "v = √ ( (2q ∆V)/m))"
then I replaced in "r = m v / (| q |B)", v with "√ ( (2q ∆V)/m))", and found out that R = (2√(2)) r.
Then for the second point,
I imposed Lorenz Force Law = Centripetal force and isolated v.
Then I substituted v in K
and obtained that "K= (1/2 ) F r".
Is this correct?
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