Field strength to deflect electron in a cathode-ray tube

[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?

Homework Equations

[/B]
Δx=vt+0.5aΔt²
KE=0.5mv²
U=qΔV
F=qvBsin(θ)
F=ma
B=ma/qvsin(θ)
K_i+U_i=K_f+U_f

The Attempt at a Solution

[/B]
Since K_i=0, then K_f=-ΔU
Since U=qΔV, then K_f=-qΔV
Since K_f=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=v_iΔt+0.5a_xΔt²
Since a_x=0, because the force should be acting to change only the a_y
t=4.047×10^-10 s
so, Δy=v_iΔt+0.5a_yΔt²

and since Δy=0.024tan(13)=5.54×10^-3

then, a_y=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!

 

[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.

 

[David23454]

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?

 

[kuruman]

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.

 

[David23454]

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

a_c=V²/r

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

 

[kuruman]

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

 

[David23454]

Ye

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.

 

[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}.

 

[David23454]

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

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

 

[kuruman]

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.

 

[David23454]

ok

 

[kuruman]

Please consider post #8.

 

[David23454]

x(t)=v_x+0.5a_xt² and y(t)=v_yt+0.5a_yt²

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.

 

[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?

 

[David23454]

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

 

[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?

 

[haruspex]

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?