- #1
Antoha1
- 14
- 2
- Homework Statement
- A particle with a charge equal to that of an electron enters a magnetic field. The direction of the particle's velocity
makes an angle of 45 degrees with the magnetic field lines. The particle moves in a spiral with a pitch of
##l##=2 cm. What is the magnitude of the particle's momentum when the magnetic field induction B is 0.02 T ?
- Relevant Equations
- ##F_{Lorenz}=qvBsin\alpha##
##F_{centrifugal}=\frac{mv^{2}}{R}##
In this case:
##F_{Lorenz}=F_{centrifugal}##
to calculate momentum (p), do I need to use sum of speed vectors? Maybe someone can help me to solve this problem. For now, my solution looks like this:
$$a_{centrifugal}=\frac{v^2}{R}=\frac{F_{Lorenz}}{m}=\frac{qvBsin\alpha}{m}$$
$$\frac{v}{R}=\frac{qBsin\alpha}{m}\implies m=\frac{qRBsin\alpha}{v}$$
then getting R (T here is time period):
while particle is spinning around magnetic field lines (around x) and going to the side (direction x), the circumstance and ##l## are done in the same period of time (T)
$$T_{1}=T_{2}$$
$$\frac{l}{v_{x}}=\frac{2\pi R}{v}$$
$$v_{x}=vcos\alpha$$
Then R:
$$R=\frac{l}{2\pi cos\alpha}$$
then getting back and putting R there:
$$p=mv_{x}=mvcos\alpha=\frac{qBRsin\alpha}{v}\cdot vcos\alpha=\frac{qBlsin\alpha}{2\pi cos\alpha}\cdot cos\alpha=\frac{Blqsin\alpha}{2\pi}$$
$$p=\frac{Blqsin\alpha}{2\pi}$$
I am not sure about momentum being calculated like that, should it be sum of both vectors maybe, or how it works? can someone explain to me, or solve this problem? Thanks.
$$a_{centrifugal}=\frac{v^2}{R}=\frac{F_{Lorenz}}{m}=\frac{qvBsin\alpha}{m}$$
$$\frac{v}{R}=\frac{qBsin\alpha}{m}\implies m=\frac{qRBsin\alpha}{v}$$
then getting R (T here is time period):
while particle is spinning around magnetic field lines (around x) and going to the side (direction x), the circumstance and ##l## are done in the same period of time (T)
$$T_{1}=T_{2}$$
$$\frac{l}{v_{x}}=\frac{2\pi R}{v}$$
$$v_{x}=vcos\alpha$$
Then R:
$$R=\frac{l}{2\pi cos\alpha}$$
then getting back and putting R there:
$$p=mv_{x}=mvcos\alpha=\frac{qBRsin\alpha}{v}\cdot vcos\alpha=\frac{qBlsin\alpha}{2\pi cos\alpha}\cdot cos\alpha=\frac{Blqsin\alpha}{2\pi}$$
$$p=\frac{Blqsin\alpha}{2\pi}$$
I am not sure about momentum being calculated like that, should it be sum of both vectors maybe, or how it works? can someone explain to me, or solve this problem? Thanks.
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