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Potatochip911
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Homework Statement
A source injects an electron of speed ##v## into a uniform magnetic field of magnitude ##B##. The velocity of the electron makes an angle ##\theta## with the direction of the magnetic field. Find the distance ##d## from the point of injection at which the electron next crosses the field that passes through the injection point.
Homework Equations
##F_B=m\frac{v^2}{r}##
##T=\frac{2\pi r}{v}##
The Attempt at a Solution
We want to find how much vertical displacement there is in one period essentially. To find ##T## in terms on known quantities:
$$
F_B=q(v\times B)=qvB\sin\theta=\frac{v^2}{r}\Longrightarrow r=\frac{mv}{qB\sin\theta} \\
T=\frac{2\pi r}{v}=\frac{2\pi m}{qB\sin\theta}
$$
Now the vertical displacement would be given by ##v\cdot\cos\theta## multiplied by the time (period in this case) Therefore:
$$
d=v\cos\theta\cdot \frac{2\pi m}{qB\sin\theta}=\frac{2\pi m v \mbox{ cotan }\theta}{qB}
$$
In my solution manual however they have completely neglected ##F_B=q(v\times B)## and just used the result for when ##\theta=90## (##F_B=qvB##), in the textbook the angle given is not equal to 90 so this isn't the case. Is there something I'm missing here or did they solve it incorrectly?
Edit: Some errors in my derivation:
$$F_B=qvB\sin\theta=m\frac{v^2}{r}$$
the velocity ##v^2/r## in this term is the velocity that is perpendicular to the force so it becomes: ##qvB\sin\theta=m\frac{(v\sin\theta)^2}{r}## This solves the problem.
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