Cathode Ray Tube with magnetic field

[bob dobilina]

Homework Statement

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/OLIRC3j2. Relevant equation
E=Energy
m=mass
v=velocity
q=charge
V=potential difference
d=distance
a=acceleration

V=E/q
E=(1/2)mv²
v(final)²=v(initial)² + 2ad

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

 

[haruspex]

Homework Statement

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/OLIRC3j2. Relevant equation
E=Energy
m=mass
v=velocity
q=charge
V=potential difference
d=distance
a=acceleration

V=E/q
E=(1/2)mv²
v(final)²=v(initial)² + 2ad

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

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.

 

[bob dobilina]

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.

v(final)2=v(initial)2 + 2ad

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
E = (1/2)m(2ad)²
Therefore, potential difference should be
V=E/q
V=((1/2)m(2ad)²)/q

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

 

[haruspex]

v(final)2=v(initial)2 + 2ad

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

 

[bob dobilina]

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

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

 

[haruspex]

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

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

 

[bob dobilina]

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

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

 

[haruspex]

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

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?

 

[bob dobilina]

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?

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?

 

[haruspex]

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?

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

 

[bob dobilina]

Ok, thank you.

 

[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.

 

[rude man]

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.

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?

 

[bob dobilina]

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?

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