Chapter 28
(# 6, 7, 11, 16, 26, 33, 79, 84)
6. An electric field of 1.50 kV/m and a magnetic field of 0.400 T act on a moving
electron to produce no net force. If the fields are perpendicular to each other, what
is the electron’s speed?

  
6. Letting F  q E  v  B  0 , we get vB sin  = E. We note that (for given values of
d
i
the fields) this gives a minimum value for speed whenever the sin  factor is at its
maximum value (which is 1, corresponding to  = 90°). So
vmin = E / B = (1.50  103 V / m) / (0.400 T) = 3.75  103 m / s.
7. In Fig. 28-33, an electron accelerated from rest through potential difference V1 =
1.00 kV enters the gap between two parallel plates having separation d = 20.0 mm
and potential. Neglect fringing and assume that the electron’s velocity vector is
perpendicular to the electric field vector between the plates. In unit-vector notation,
what uniform magnetic field allows the electron to travel in a straight line in the gap?
7. Straight line motion will result from zero net force acting on the system; we ignore

  
 
 
gravity. Thus, F  q E  v  B  0 . Note that v B so v  B  vB . Thus, obtaining the
d
i
speed from the formula for kinetic energy, we obtain
100 V  20 103m 
E
E
B 

 2.67 104 T.
3

19

31
v
2 K / me
2 1.0 10 V  1.60 10 C  /  9.1110 kg 
In unit-vector notation, B  (2.67 104 T)kˆ .
11. A metal strip 6.50 cm long, 0.850 cm wide, and 0.760 mm thick moves with
constant velocity v through a uniform magnetic field B = 1.20 mT directed
perpendicular to the strip, as shown in Fig. 28-35. A potential difference of 3.90 µV
is measured between points x and y across the strip. Calculate the speed v.

  

11. For a free charge q inside the metal strip with velocity v we have F  q E  v  B .
d
i
We set this force equal to zero and use the relation between (uniform) electric field and
potential difference. Thus,
c
h
3.90  109 V
E Vx  Vy d xy
v 

 0.382 m s .
B
B
120
.  103 T 0.850  102 m
c
1135
hc
h
1136
CHAPTER 28
16. An electron is accelerated from rest by a potential difference of 350 V. It then
enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular
to the field. Calculate (a) the speed of the electron and (b) the radius of its path in
the magnetic field.
16. (a) The accelerating process may be seen as a conversion of potential energy eV into
1
kinetic energy. Since it starts from rest, me v 2  eV and
2


2 1.60 1019 C  350 V 
2eV
v

 1.11 107 m s.
me
9.111031 kg
(b) Eq. 28-16 gives




9.111031 kg 1.11107 m s
me v
r

 3.16 104 m.
19
3
eB
1.60 10 C 200 10 T


26. In Fig. 28-38, an electron with an initial kinetic energy of 4.0 keV enters region 1
at time t = 0. That region contains a uniform magnetic field directed into the page,
with magnitude 0.010 T. The electron goes through a half-circle and then exits
region 1, headed toward region 2 across a gap of 25.0 cm. There is an electric
potential difference ΔV = 2000 V across the gap, with a polarity such that the
electron’s speed increases uniformly as it traverses the gap. Region 2 contains a
uniform magnetic field directed out of the page, with magnitude 0.020 T. The
electron goes through a half-circle and then leaves region 2. At what time t does it
leave?
26. Eq. 28-17 gives T = 2me /eB. Thus, the total time is
1
T 
 T  me  1
 2  + tgap +  2  = e B + B  + tgap .
 1
 2
 1
2
The time spent in the gap (which is where the electron is accelerating in accordance with
Eq. 2-15) requires a few steps to figure out: letting t = tgap then we want to solve
1
d = vo t + 2 a t2
0.25 m =
2Ko
1 e V 2
me t + 2 me d t
1137
for t. We find in this way that the time spent in the gap is t  6 ns. Thus, the total time is
8.7 ns.
33. A horizontal power line carries a current of 5000 A from south to north. Earth’s
magnetic field (60.0 µT) is directed toward the north and inclined downward at
70.0º to the horizontal. Find the (a) magnitude and (b) direction of the magnetic
force on 100 m of the line due to Earth’s field.
33. (a) The magnitude of the magnetic force on the wire is given by FB = iLB sin ,
where i is the current in the wire, L is the length of the wire, B is the magnitude of the
magnetic field, and  is the angle between the current and the field. In this case  = 70°.
Thus,
b
gb gc
h
FB  5000 A 100 m 60.0  106 T sin 70  28.2 N .

 
(b) We apply the right-hand rule to the vector product FB  iL  B to show that the force
is to the west.
79. A proton, a deuteron (q = +e, m = 2.0 u), and an alpha particle (q = +2e, m = 4.0 u)
are accelerated through the same potential difference and then enter the same
region of uniform magnetic field B, moving perpendicular to B. What is the ratio of
(a) the proton’s kinetic energy Kp to the alpha particle’s kinetic energy Kα and (b)
the deuteron’s kinetic energy Kd to Kα? If the radius of the proton’s circular path is
10 cm, what is the radius of (c) the deuteron’s path and (d) the alpha particle’s path?
79. (a) Since K = qV we have K p  12 K  as q  2 K p  , or K p / K  0.50.
(b) Similarly, q  2 K d , K d / K  0.50.
(c) Since r  2mK qB  mK q , we have
rd 
md K d q p rp

m p K p qd
 2.00u  K p
r  10
1.00u  K p p
2cm=14cm.
(d) Similarly, for the alpha particle, we have
r 
m K q p rp

mp K p q
 4.00u  K erp  10
1.00u   K 2  2e
2cm=14cm.
1138
CHAPTER 28
84. A proton, a deuteron (q = +e, m = 2.0 u), and an alpha particle (q = +2e, m = 4.0 u)
all having the same kinetic energy enter a region of uniform magnetic field B,
moving perpendicular to B. What is the ratio of (a) the radius rd of the deuteron
path to the radius rp of the proton path and (b) the radius rα of the alpha particle
path to rp?
84. Referring to the solution of problem 19 part (b), we see that r  2mK qB implies
the proportionality: r  mK qB . Thus,
(a)
rd
md K d q p
2.0u e


 2  1.4 , and
rp
m p K p qd
1.0u e
(b)
r
m K q p
4.0u e


 1.0.
rp
m p K p q
1.0u 2e