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Homework Statement
An electron moves in a straight line at a speed of 6.0 \times 10^7m/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
F=qv \times B
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 F=qvB sin(\theta) 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). B=\frac{\mu_0 I}{2 \pi r}
=\frac{(4\pi \times 10^{-7})I}{2\pi (0.015)} =1.33 \times 10^{-5}IIt'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!