Deflection of electron beam in cathode ray tube

[skate_nerd]

Homework Statement

Part a) Deflection of the electron beam in a cathode ray tube television set has electrons accelerated until they have K.E. of 25 keV. What is the magnetic field magnitude required to deflect the electron beam 3.3 cm when the magnetic field is 8 cm long?
Part b) When there is a gap of 22 cm between the end of the magnetic field and the end of the CRT, what is the total deflection of the beam? In other words, how far from the center of the screen does the beam hit?

Homework Equations

F=ma, F=qvB, K.E.=.5mv², d=vt, d=v₀t+.5at², v_f²=v₀²+2ad.

The Attempt at a Solution

Pretty long problem, I got through this whole thing, parts a and b, went through my work several times, and still have a slightly different answer than what my teacher says is correct.

Part a) First off, using K.E.=.5mv², I found the velocity in the horizontal or x direction:

.5mv_x²=25 keV
v_x=√(2(25keV)/m) where m=mass of an electron=9.109(10^-31). so:
v_x=9.383(10⁷) m/s

Next, found the time needed to reach the end of the plates' magnetic field:

t=d/v_x where d=length of plates=8 cm=.08 m
t=(.08m)/(9.383(10⁷)m/s)
t=8.526(10^-10) s

Next, found the vertical or y direction acceleration component to achieve the 3.3 cm displacement by the end of the plates' field:

d_y=v_0yt+.5a_yt² solved for a_y and v_0y=0 gives
a_y=2d_y/t²
a_y=2(.033m)/(8.526(10^-10)s)²
a_y=9.08(10¹⁶) m/s²

Now to find the final velocity in the vertical or y-direction:

v_fy²=v_0y²+2a_yd_y solved for v_fy where v_0y=0
v_fy=√(2(a_yd_y))
v_fy=√(2(9.08(10¹⁶)m/s²)(0.033m))
v_fy=7.74(10⁷) m/s

Finally, using F=ma & F=qvB:

ma=qvB, and then solving for B gives
B=ma/qv, using v_fy and a_y and also mass of an electron=9.109(10^-31) kg and charge of an electron=1.602(10^-19) C:
B=[(9.109(10^-31)kg)(9.08(10¹⁶)m/s²)]/[(1.602(10^-19)C)(7.74(10⁷)m/s)
B=6.669(10^-3) Teslas.

Teacher's answer: 4.699(10^-3) Teslas.

Part b) Used v_x=d_x/t solved for t to find the time to get 22 cm to the screen:

t=(0.22m)/(9.383(10⁷)m/s)
t=2.345(10^-9) s

Since this time is the same for the vertical or y direction, I used the same formula solved for d_y to get the vertical deflection:

d_y=(7.74(10⁷)m/s)(2.345(10^-9)s)
d_y=0.1815 meters.

Have to add the initial deflection in between the plates' magnetic field to get the total vertical deflection:

0.1815m+0.033m=0.2145 meters or 21.45 cm.
Teacher's answer: 25.16 cm.

So seeing as I also got the second answer wrong which only uses v_y and a_y, that's telling me I went wrong in my process somewhere up until finding v_y.

If anybody could help point out where I messed up that would be great. Thanks!

 

[haruspex]

The acceleration due to a magnetic field is at right angles to the direction of travel, right? So the trajectory is an arc of a circle. For small deflections over long paths that doesn't matter much, but here the deflection is 40% of the path length.
Calculating it that way (and trusting all else of your calculations) I get 5.7E-3 Teslas.

 

[skate_nerd]

That all sounds pretty accurate but I'm pretty sure that isn't how we are to solve it because we've never learned anything like that.
Also from however you did it I don't see any easy way to find the total deflection at the end.

 

[haruspex]

Also from however you did it I don't see any easy way to find the total deflection at the end.

You're told the deflection. Do you mean, how to find the radius of curvature? Just a bit of geometry. If it turns through an angle θ of an arc radius r then you have distance to target T = r sin(θ) and deflection d = r(1-cos(θ)).
Anyway, as I said, I had not checked the rest of your calculation. Now that I do so, I see you have used v_y as the v in the B=ma/qv equation. That's wrong. Since you're treating it as purely a sideways deflection, the acceleration depends on v_x, not v_y. Correcting that, but using your method, I get 5.5E-3.
Because I relied on your calculation for that part, I picked up the same error. Correcting that with my method I get 4.71E-3. So it seems your teacher is using my method.

 

[skate_nerd]

I do see what you mean about the curving position, in between the plates. I actually can do that. However for the total deflection, part B, it says if there is an addition length of 22 cm past the magnetic plates, then where would the beam hit the screen. Past the plates there would be no more acceleration, just the two components of velocity it has after the plates, so it would move in a straight light to a certain displacement from the center of the screen 22 cm ahead. But I think I could do that now. Thanks for the help on the arc length part though.