Field region beams penetrate

rude man

1 Tesla = 10,000 Gauss. Your teach is trying to make life even more miserable than it has to be ... tell him you won't do any more problems in any other than SI units...

Actually, pure physicists tend to use the cgs system which includes the Gauss. Applied physicists and engineers gravitate towards the SI system, aka 'rationalized mks" system ("meter, kilogram, second").
.

warnexus

he did teach us how to convert units super fast in the first physics lecture. he really want us to convert units in some assigned problems. but it's for practice which I totally needed but am getting better at. for this one it would be :

90 Gauss * (1 Tesla/10,000 Gauss) = .009 Tesla.

okay now to plug these values in:

charge * velocity * magnetic field = (mass of an electron) (2nd derivative of position respect to time)

(1.6 * 10 ^ -9 C) * (8.4 * 10 ^ 6 m/s) (.009 T) = (9.1 * 10 ^ -31 kg) (2nd derivative of position respect to time.

well im left with acceleration given as a constant value of 1.3 * 10 ^ 26 m/s^2 . I could integrate this twice if this was a function so I can get distance but its just a constant value

rude man

Right!
Don't mess with numbers now. You need to get your equations first.

So why not start with the z-axis. What does F = ma look like for it? Keep in mind that the velocity is in x and the B field is in y direction, so use the appropriate velocity component ...

mfb

Right, but the deflection in z-direction is important.

It is not. There is a very simple solution if you know some basics about that type of motion involved here.

warnexus

well if we start with the z axis, the z axis is the axis that is towards you. there is no velocity in the z, so its zero

charge * velocity * magnetic field = m ( acceleration)
0 = m(acceleration)

mass * acceleration is also zero

rude man

Oh no? If the initial velocity is in the x direction and the B field is in the y direction, you don't expect to see any velocity buildup in the z direction?

Write your F = ma for the z axis.

warnexus

is that because the direction of velocity is perpendicular to the magnetic field that's why there is velocity build-up?

i will try to write one out.

(charge) (velocity)(magnetic field) = (mass of an electron) (acceleration_sub z)

rude man

That's good, but which velocity component is the appropriate one?
Remember, F = qv x B is a vector equation.

warnexus

(charge)(velocity sub_z)(magnetic field) = (mass of an electron) (acceleration sub_z)

rude man

Nope. What direction does the velocity have to be in order for the B field, which is in the y direction, to give motion in the z direction?

warnexus

velocity would need to be in the y direction

rude man

No. Velocity has to be at right angles to the B field and to the z direction in order to produce motion along the z direction, right? And don't write the full names, use q for charge, B for mag field, v for velocity ...

BTW do you use vectors in your physics course?

warnexus

Yes, Vectors was covered... Oh I see now! Kinematic Equations! Now I know.

rude man

OK, so back to writing the equation for the z motion ...

warnexus

q v sin(90) B = m * a..