Field strength to deflect electron in a cathode-ray tube

• David23454
In summary, the magnetic force is perpendicular to the velocity, and the angle that you seeking has to do with the circular arc of the electron's trajectory.
David23454

Homework Statement

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?

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

The Attempt at a Solution

[/B]
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!

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.

kuruman said:
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.

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?

David23454 said:
Since ax=0, because the force should be acting to change only the ay
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.

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?

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

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

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:
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

David23454 said:
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
You are going around in a circle. This gives you B = B if you simplify.

ok

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.

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?

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:
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?

David23454 said:
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?
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 traveled along the x axis?

1. What is field strength in a cathode-ray tube?

Field strength in a cathode-ray tube refers to the strength of the electric field created by the charged plates inside the tube that deflect electrons.

2. How is field strength related to electron deflection in a cathode-ray tube?

The strength of the electric field determines the amount of force applied to the electrons, which in turn affects the degree of deflection in the cathode-ray tube.

3. What factors affect field strength in a cathode-ray tube?

The distance between the charged plates, the voltage applied to them, and the charge of the particles being deflected can all influence the field strength in a cathode-ray tube.

4. How can field strength be calculated in a cathode-ray tube?

The field strength can be calculated by dividing the voltage by the distance between the charged plates, or by using the equation F = qE, where F is the force, q is the charge, and E is the electric field.

5. Why is field strength important in a cathode-ray tube?

The field strength is crucial in a cathode-ray tube as it controls the deflection of the electrons, which ultimately determines the pattern and image produced on the screen.

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