# Field strength to deflect electron in a cathode-ray tube

1. May 20, 2017

### David23454

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

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 2.4-cm-wide region of uniform magnetic field in the figure(Figure 1) .

What field strength will deflect the electron by θ = 13 degrees?

2. Relevant equations

Δx=vt+0.5aΔt2
KE=0.5mv2
U=qΔV
F=qvBsin(θ)
F=ma
B=ma/qvsin(θ)
Ki+Ui=Kf+Uf

3. The attempt at a solution

Since Ki=0, then Kf=-ΔU
Since U=qΔV, then Kf=-qΔV
Since Kf=0.5mv2, then 0.5mv2=-qΔV

Using 0.5mv2=-qΔV, and inputing values 0.5(9.11×10-31)v2)=-(-1.60×10-19)(10,000)

So, v after acceleration by ΔV is 5.931×107 m/s.

Δx=viΔt+0.5axΔt2
Since ax=0, because the force should be acting to change only the ay
t=4.047×10-10 s
so, Δy=viΔt+0.5ayΔt2

and since Δy=0.024tan(13)=5.54×10-3

then, ay=6.765×1016

Since: F=qvBsin(θ) and F=ma, then
B=ma/qvsin(θ), and θ=90, (I think?)
so
B=(9.11×10-31)(6.756×1016)/(1.60×10-19)(5.931×107)=6.49×10-3 T

This answer is incorrect. Could someone show me where I went wrong? Thanks!

2. May 20, 2017

### kuruman

The magnetic force is perpendicular to the velocity. In F = q v B sinθ, θ = π/2. The angle that you seeking has to do with the circular arc of the electron's trajectory.

3. May 20, 2017

### David23454

Right. I'm pretty sure I used π/2=90 degrees in my calculations. I'm trying to find the field strength needed to deflect the electron. Any idea where I went wrong in my calculations?

4. May 20, 2017

### kuruman

This is where you went wrong. What I am trying to say is that the electron undergoes uniform circular motion in he region of magnetic field. The acceleration is centripetal.

5. May 20, 2017

### David23454

Oh, I think I see. So, the acceleration I should use to input into the equation B=ma/qv should be:

ac=V2/r

How would I calculate the "r" in the equation for centripetal acceleration?

6. May 20, 2017

### kuruman

The "r" depends on the magnetic field. Can you find an expression for it in terms of the magnetic field?

7. May 20, 2017

### David23454

Ye
Yes, r=mv/qB, but since the magnetic field is what I'm trying to find, I'm not sure what I can do with it.

8. May 20, 2017

### kuruman

Remember that the electron is undergoing uniform circular motion. Can you write expressions for the x and y components of its position as a function of time? Note that at t = 0, the electron is at position {0, mv/qB} and moving with velocity {v, 0}.

Last edited: May 20, 2017
9. May 20, 2017

### David23454

Wait, I think I might be getting it, so the equation would look like :

since ac=v2/(mv/qB) then
B=m((v2)/(mv/qB))/qv

10. May 20, 2017

### kuruman

You are going around in a circle. This gives you B = B if you simplify.

11. May 20, 2017

### David23454

ok

12. May 20, 2017

### kuruman

13. May 20, 2017

### David23454

x(t)=vx+0.5axt2 and y(t)=vyt+0.5ayt2

I know that I have to incorporate something relating to the fact that at t=0 where position is (0, mv/qB) and velocity is (v,0), but I'm not sure how to proceed.

14. May 20, 2017

### kuruman

Please forget the equations that you quoted. They are not pertinent to uniform circular motion. In uniform circular motion the object is going around a circle at constant speed. This would be the case here if the magnetic field extended over a larger region. Instead the particle exits the field, nevertheless while it's still in the field region, it moves at constant speed.
Try this approach. Draw a picture of the particle at the exit point and a vector from the center of the circle (remember it has radius mv/qB). The x-component of the position vector must be what?

15. May 20, 2017

### David23454

The x-component of the position should be the length of box containing the magnetic field (xf-xi)=0.024 m, and the y-component of the position should be y=(0.024)tan(13).

Last edited: May 20, 2017
16. May 21, 2017

### David23454

Using geometry, I think the radius of the should be (Δx2+Δy2)/2Δy, which should be 0.05476 m. Then I should be able to put it in the formula B=mv/rq. Does this sound correct?

17. May 21, 2017

### haruspex

No.
Draw a diagram showing the arc of travel and the centre of the arc, joining the centre to the ends of the arc with two radii to form a sector.
What is the angle in the sector, i.e. between the radii?
In terms of that angle and the radius, what is the distance travelled along the x axis?