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## Field region beams penetrate

 Quote by warnexus oh i see. interesting. btw how come magnetic field is given in Gauss, the previous questions that involved with magnetic field were given in Tesla? Wouldn't Tesla have to be used?
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").
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 Quote by 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"). .
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

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 Quote by 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)
Right!
 (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
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 ...

Mentor
 Quote by rude man Request not granted . And the answer is in x, not z.
Right, but the deflection in z-direction is important.

 Quote by rude man This is a pretty sophisticated problem, involving coupled motions.
It is not. There is a very simple solution if you know some basics about that type of motion involved here.

 Quote by 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 ...
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

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 Quote by 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
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.

 Quote by 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.
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)

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

 Quote by rude man That's good, but which velocity component is the appropriate one? Remember, F = qv x B is a vector equation.
(charge)(velocity sub_z)(magnetic field) = (mass of an electron) (acceleration sub_z)

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 Quote by warnexus (charge)(velocity sub_z)(magnetic field) = (mass of an electron) (acceleration sub_z)
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?

 Quote by 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?
velocity would need to be in the y direction

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 Quote by warnexus velocity would need to be in the y direction
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?

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

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 Quote by warnexus Yes, Vectors was covered... Oh I see now! Kinematic Equations! Now I know.
OK, so back to writing the equation for the z motion ...

 Quote by rude man OK, so back to writing the equation for the z motion ...
q v sin(90) B = m * a..

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