Download Problem 20.28 Since the electrons start from rest and

Problem 20.28
Since the electrons start from rest and speed up, plate B must have a positive charge.
The electrons experience an electric force downward, indicating that the electric field between the deflection plates points
upward. Note that no force is required to keep the electrons moving in the horizontal direction at constant speed; since
the tube is evacuated they do not collide with anything.
The magnetic field points into the page, so the magnetic force – ev × B on an electron must be upward.
When the electron passes through the accelerating plates:
∆K + ∆U = 0
∆K = – ∆U
K f – 0 = – ( – e ) ∆V acc
1 2
--- mv f = e∆V acc
2
vf =
2e ( ∆V acc )
-----------------------m
If the accelerating potential difference is 5 kilovolts, then
– 19
vf =
3
2 ( 1.6 ×10 C ) ( 5 ×10 V )- = 4.2 ×107 m
---------------------------------------------------------------–31
s
( 9 ×10 kg )
This is fast, but not so fast that it is necessary to do a relativistically correct calculation.
At the location of the electron beam, the magnetic field due to the two coils is:
µ 0 2πR 2 I
B = ( 2 )N ------ ---------------------------4π ( z 2 + R 2 ) 3 ⁄ 2
If each coil has 300 turns, and a radius of 7 cm, the center of the coil is 3.5 cm from the beam, and the current through the
coils is 0.7A, then the magnetic field at the location of the electron beam is:
2
–7 Tm
2π ( 0.07m ) ( 0.7A ) –3
= 2.7 ×10 T
B = ( 2 ) ( 300 ) ⎛ 1 ×10 ---------⎞ ---------------------------------------------------------⎝
A ⎠ ( 0.035m 2 + 0.07m 2 )3 ⁄ 2
When the magnetic field is present, the electrons travel in a straight line at constant speed, so the downward electric force
on an electron due to the electric field between the deflection plates must be equal in magnitude and opposite in direction
to the upward magnetic force on the electron. So
F el = F mag
eE = evB sin ( 90° )
E = vB ( 1 )
7m
–3
5V
E = ⎛ 4.2 ×10 -----⎞ ( 2.7 ×10 T ) = 1.13 ×10 ----⎝
⎠
s
m
If the separation between the deflection plates is 6 mm, then
5V
∆V def = – ⎛ 1.13 ×10 -----⎞ ( 0.006m ) = 680V
⎝
m⎠