# Cathode Ray Tube with magnetic field

1. Aug 31, 2015

### bob dobilina

1. The problem statement, all variables and given/known data

A solenoid is placed beneath a CRT that produces a magnetic field of 1.10 x -2. The CRT is 1 meter away from a screen.
The electrons that make up the beam were accelerated from rest through a potential difference(V) at the beginning of the CRT. Acceleration of the electrons in the magnetic field is 9.08 x 1014 m/s2.
What is the potential difference required to accelerate the electrons.

See image for diagram http://imgur.com/OLIRC3j

2. Relevant equation
E=Energy
m=mass
v=velocity
q=charge
V=potential difference
d=distance
a=acceleration

V=E/q
E=(1/2)mv2

3. The attempt at a solution
I figure that I can solve for v(final) because I have all the variables. I can then use v(final) in the E=(1/2)mv2 formula to solve for E. Once I have E, i should be able to plug it into V=E/q.

Any help would be appreciated

2. Aug 31, 2015

### haruspex

Yes, that sounds like a plan, though you did not explain how you will find the velocity.
My advice is to solve it entirely symbolically, not plugging in any numbers until you have a final formula. You may find there's some cancellation.

3. Aug 31, 2015

### bob dobilina

Initial velocity is 0 because it is at rest, so final velocity will be equal to whatever "2ad" turns out to be.
So E will look something like this
Therefore, potential difference should be
V=E/q

I am unsure if it is okay if I use final velocity in my final equation, or if i should use average velocity?

4. Aug 31, 2015

### haruspex

I don't think so. Where are you getting a and d from for this?

5. Aug 31, 2015

### bob dobilina

a and d are given to us in the question. This formula is one of the kinematics formula provided to me.
V final 2 = Velocity initial 2 + 2ad

6. Aug 31, 2015

### haruspex

Be specific. What are you using for a and d, out of the supplied information?

7. Aug 31, 2015

### bob dobilina

For d I am using the distance of the screen from the CRT, which is 1 meter.
For a, I am using the acceleration of the electrons in the magnetic field, which is 9.08 x 1014 m/s^2

8. Aug 31, 2015

### haruspex

In the equation v2=2ad, the acceleration must be acting along the distance d. Where is the magnetic field act in relation to the 1m, and in what direction?

Last edited: Aug 31, 2015
9. Aug 31, 2015

### bob dobilina

The magnetic field is acting in a clockwise direction, which is why it pushes the electron the right side of the screen.

Hmmm, I know that E=Vq, therefore Vq=(1/2)mv2 so i need to find v.

Since the problem tells me the acceleration in the magnetic field, would I be able to use Force=massxacceleration to find the magnetic force?
Then with the magnetic force, use this formula:
Magnetic Force =qvB
(B=Magnetic Field)
,to solve for v?

10. Aug 31, 2015

### haruspex

Yes, that's how you need to find v here.

11. Aug 31, 2015

### bob dobilina

Ok, thank you.

12. Aug 31, 2015

### bob dobilina

Hi haruspex. Second part of the question here. Would appreciate any advice on this one. The only thing I can come up with is reversing the current of solenoid, but not sure how to prove this with the formulas.

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13. Aug 31, 2015

### rude man

Since haruspex may still be in bed (Australia), can I suggest (1) what effect is there on the mag. force when B is reversed, and (2) does the magnitude of the deflection depend on the magnitude of B?

14. Aug 31, 2015

### bob dobilina

Thanks rude man, I should be able to figure it out from here...