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Homework Statement
A proton (q = 1.6×10^{−19} C, m = 1.67×10^{−27} kg) moves at 300 ms^{−1} along a line midway between the x and z axes. There is a magnetic field of 1T along the x axis and an electric field of 2000 Vm^{−1} along the y axis. What is the force on the proton and what is its acceleration?
Homework Equations
Scalar product: a·b=abcosθ=a_{x}b_{x} +a_{y}b_{y} +a_{z}b_{z}.
Vector product: a×b = i(a_{y}b_{z} −a_{z}b_{y})+j(a_{z}b_{x} −a_{x}b_{z})+k(a_{x}b_{y} −a_{y}b_{x}), a×b = absinθ.
Forces on a charged particle in electric and magnetic fields: F = qE + qv × B.
The Attempt at a Solution
First off, my main assumption is that there shouldn't be any relativity involved in this question since v=300m s^{1} is not particularly fast. However when I arrived at the answer I have calculated below, that is quite a lot of acceleration! I didn't end up needing to use the scalar and vector product formulae that were provided on the question sheet, so I fear that I am missing something!
I have drawn a diagram to show how I have interpreted the question, which is attached. The proton is traveling with initial velocity v=300m s^{1} along a line where x=z and y=0 (which I assume is what my lecturer means when he says "along a line midway between the x and z axes).
So the total force acting on the particle F=F_{E}+F_{B} where F_{E} is the force due to the electric field of 2000 Vm^{−1} and F_{B} is the force on the proton due to the magnetic field of 1 T.
So F=qE+qvB
But what is v? Because only the component of motion perpendicular to the magnetic field (the velocity along the zaxis) affects the force, v=300sin(Pi/4)=[tex]150\sqrt{2}[/tex] m s^{1}.
Using the right hand slap rule for the magnetic field, I can see that the force exerted on the proton by the magnetic field is in the same direction as that exerted by the electric field on a positive particle, that is, straight along the yaxis. Therefore I should be able to just add these two forces.
Therefore F = 1.6×10^{−19} C × 2000 Vm^{1} + 1.6×10^{−19} C × [tex]150\sqrt{2}[/tex] m s^{1} × 1 T = 3.54×10^{−16} N along the yaxis.
Then using a=F/m = (3.54×10^{−16} N) / (1.67×10^{−27} kg) = 2.12×10^{11} m s^{2} in the ydirection.
Does all of this working seem to make sense to you? I can't shake the feeling that I have missed something  why would my lecturer give me the scalar and vector product formulae if I did not have to use them?
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