(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A proton (q = 1.6×10^{−19}C, m = 1.67×10^{−27}kg) moves at 300 ms^{−1}along a line midway between thexandzaxes. There is a magnetic field of 1T along thexaxis and an electric field of 2000 Vm^{−1}along theyaxis. What is the force on the proton and what is its acceleration?

2. Relevant 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.

3. The attempt at a solution

First off, my main assumption is that there shouldn't be any relativity involved in this question sincev=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 velocityv=300m s^{-1}along a line wherex=zandy=0 (which I assume is what my lecturer means when he says "along a line midway between thexandzaxes).

So the total force acting on the particleF=F+_{E}Fwhere_{B}Fis the force due to the electric field of 2000 Vm_{E}^{−1}andFis the force on the proton due to the magnetic field of 1 T._{B}

SoF=qE+qvB

But what isv? Because only the component of motion perpendicular to the magnetic field (the velocity along thez-axis) 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 they-axis. Therefore I should be able to just add these two forces.

ThereforeF= 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 they-axis.

Then using a=F/m = (3.54×10^{−16}N) / (1.67×10^{−27}kg) = 2.12×10^{11}m s^{-2}in they-direction.

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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# Homework Help: A particle accelerating in both magnetic and electric fields

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