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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**=

*ab*cos

*θ*=

*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**| =

*ab*sin

*θ*.

Forces on a charged particle in electric and magnetic fields:

**F**=

*q*

**E**+

*q*

**v**×

**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**where

_{B}**F**is the force due to the electric field of 2000 Vm

_{E}^{−1}and

**F**is the force on the proton due to the magnetic field of 1 T.

_{B}So

**F**=

*q*

**E**+

*q*

**v**

**B**

But what is

**v**? Because only the component of motion perpendicular to the magnetic field (the velocity along the

*z*-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 the

*y*-axis. 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

*y*-axis.

Then using a=

**F**/m = (3.54×10

^{−16}N) / (1.67×10

^{−27}kg) = 2.12×10

^{11}m s

^{-2}in the

*y*-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?