- #1

- 62

- 0

## 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.=.5mv

^{2}, d=vt, d=v

_{0}t+.5at

^{2}, v

_{f}

^{2}=v

_{0}

^{2}+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.=.5mv

^{2}, I found the velocity in the horizontal or x direction:

.5mv

_{x}

^{2}=25 keV

v

_{x}=√(2(25keV)/m) where m=mass of an electron=9.109(10

^{-31}). so:

v

_{x}=9.383(10

^{7}) 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(10

^{7})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

_{0y}t+.5a

_{y}t

^{2}solved for a

_{y}and v

_{0y}=0 gives

a

_{y}=2d

_{y}/t

^{2}

a

_{y}=2(.033m)/(8.526(10

^{-10})s)

^{2}

a

_{y}=9.08(10

^{16}) m/s

^{2}

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

v

_{fy}

^{2}=v

_{0y}

^{2}+2a

_{y}d

_{y}solved for v

_{fy}where v

_{0y}=0

v

_{fy}=√(2(a

_{y}d

_{y}))

v

_{fy}=√(2(9.08(10

^{16})m/s

^{2})(0.033m))

v

_{fy}=7.74(10

^{7}) 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(10

^{16})m/s

^{2})]/[(1.602(10

^{-19})C)(7.74(10

^{7})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(10

^{7})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(10

^{7})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!