- #1
Callix
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Homework Statement
An electron moves in a straight line at a speed of [itex]6.0 \times 10^7[/itex]m/s. Calculate the magnitude and direction of the magnetic field at a position 0.005m behind the electron and 0.015m below its line of motion.
Homework Equations
[itex]F=qv \times B[/itex]
The Attempt at a Solution
I'm not really even sure if I drew the scenario correctly.. I figured that since there is a flow/movement of an electron, that it will current some sort of current in that direction (well technically, in the opposite direction to that of the electron's velocity vector). So then I modeled the field as if there is a current in a wire. The way I answered the questions (which need checking...) was by using the diagram that I drew..
a). I said that [itex]F=qvB sin(\theta)[/itex] but I'm not sure what the angle is.. If the electron produces a magnetic field, won't it extend almost forever to the left? So that would mean that the angle is 180 between v and B?
b). [itex]B=\frac{\mu_0 I}{2 \pi r}[/itex]
[itex]=\frac{(4\pi \times 10^{-7})I}{2\pi (0.015)} =1.33 \times 10^{-5}I[/itex]It's all probably wrong.. If someone could give a detailed explanation as to what's going on and what how I may proceed, that would be greatly appreciated. Thanks!