Physics 213—Problem Set 10—Solutions
Fall 1997
1. Reading Assignment
a) This problem set covers Serway 29.1-29.6, 30.1-30.2
b) For next week, we will cover material in chapter 30 up to 30.7 and the beginning of chapter 31. For
the final lab (which starts 11/12) you should read a bit ahead on Faraday’s law.
2. Serway 29-58
Indicate the initial direction of the deflection of the charged particles as they enter the magnetic fields
shown in Figure P29.58 of your text.
SOLUTION: The initial direction of deflection is the direction of the force on the charge as it enters the
magnetic field.
(a) toward the top of the page
(b) out of the page
~
(c) no deflection, because ~v is antiparallel to B
(d) into the page
3. Serway 29-14
A conductor suspended by two flexible wires as in Figure P29.14 of your text has a mass per unit length
of 0.040 kg/m.
a) What current must exist in the conductor in order for the tension in the supporting wires to be zero
when the magnetic field is 3.6 T into the page?
b) What is the required direction for the current?
SOLUTION:
a)The magnetic force has to be equal to the weight of the conductor, and directed upward. Let l̂ be a unit
~ and a mass per
vector in the direction of the current flow. With a magnetic force per unit length I l̂ × B
unit length µ we get
µg
=
I
=
IB
0.040 kg/m × 9.8 N/kg
µg
N
=
= 0.11
= 0.11 A
B
3.6 T
mT
N
represents a current, as you can see from F = IlB.
To check the dimensions: The combination mT
b)The current has to flow from left to right so that the magnetic force is directed up.
4. Serway 29-23A
A rectangular loop consists of N closely wrapped turns and has dimensions a and b. The loop is hinged
along the y axis, and its plane makes an angle θ with the x axis (see figure P29.23).
a) What is the magnitude of the torque exerted on the loop by a uniform magnetic field B directed
along the x axis when the current is I in the direction shown?
b) What is the expected direction of rotation of the loop?
SOLUTION: The xz-plane will be called horizontal in this solution, and the y-direction vertical. a is the
length of the horizontal sections of the loop in Figure P29.23, and b the length of the vertical sections. ~a is
~ = B x̂.
parallel to the horizontal sections and points away from the y-axis. B
a)There is no torque due to the forces on the horizontal parts of the loop, because these forces act along
the same line. The torque about the y-axis, where the loop is hinged, is due to the magnetic force acting
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on the opposite vertical section of the loop of length b. The current in that section is directed down, in the
negative y-direction. The torque on a loop with N turns is
~
τ
=
~ = ~a × NIb(−ŷ) × B
~ = NIbB ~a × [(−ŷ) × x̂]
~a × F
=
NIbB ~a × ẑ
=
NabIB sin (
π
− θ) (−ŷ) = −NabIB cos θ ŷ
2
Note that θ is the angle between ~a and the x-axis, not the z-axis.
Alternately we could have calculated the vector product of the magnetic moment, µ = NIab, and the
~ is again ( π − θ), and their vector product points in the
magnetic field, B. The angle between ~
µ and B
2
negative y-direction; the result is the same as above.
b)The loop will rotate about the y-axis toward the z-axis, clockwise as seen from above.
5. Cornell Accelerator
CESR (the Cornell Electron Storage Ring) is housed below the new track on Alumni Field. It keeps
electrons that are traveling at almost the speed of light in circular counter-clockwise orbits with a period
of 2.5 microseconds by using a set of electromagnets.
a) What is the strength of the magnetic field needed to accomplish this, in Tesla? At such high speeds,
increases in kinetic energy are evidenced by increases in the effective mass of the particles; at CESR the
electron mass is increased by a factor of 10,000 over its rest mass.
b) How much work is done by the field on each electron in one orbit by the magnetic field?
c) What kind of orbits would be supported by these magnets for positrons, the positively charged
anti-particles of electrons.
SOLUTION:
a)For circular motion of a charge q in a magnetic field B, the period T and angular frequency ω = 2π/T
q
are related by ω = m
B. Inserting
B =
2π × 9 × 10−27 kg
2πm
=
= 0.14 T
qT
1.6 × 10−19 C 2.5 × 10−6 s
Dimensions: kg/(C s) = (kg m)/(m C s) = (N s)/(m C). Use F = qvB to see that (N s)/(m C) represents
a magnetic field.
b)Magnetic forces do no work on moving charges, because they are always perpendicular to the direction of
motion of the charges.
c)Positrons have the same orbits, circulating in the opposite direction.
6. Serway 29-49
The picture tube in a television uses magnetic deflection coils rather than electric deflection plates.
Suppose an electron beam is accelerated through a 50.0 kV potential difference and then travels through
a region of uniform magnetic field 1.00 cm wide, perpendicular to the velocity of the electrons. The
screen is located 10.0 cm from the center of the coils and is 50.0 cm wide. When the field is turned off,
the electron beam hits the center of the screen. What field strength is necessary to deflect the beam to
the side of the screen?
NOTE: The statement of the problem implies that after leaving the magnetic field the deflected beam
is directed as though it came from the center of the magnetic field region. This is not exactly true, but
it is a reasonable approximation that you should make.
SOLUTION: After acceleration through a potential difference V the electrons (charge q, mass m) have
kinetic energy K = qV = 50, 000 eV, and their speed is
r
v=
2
2qV
m
(Figure to come.) This speed remains constant until they hit the screen, at w = 25 cm above its center.
The trajectory has a circular part with radius R, and a straight line part after leaving the region of the
magnetic field. The angle of deflection from the original direction is θ which is also the angle of the arc that
the electrons travel in the magnetic field. We need to find R in order to find the necessary magnetic field
from
r
mv
2mV 1
B=
=
(6.1)
qR
q
R
In our figure, the straight line part of the trajectory is extrapolated backward to intersect with the horizontal
at the angle θ at a point which is somewhere between the center and the right edge of the magnetic field
region. The exact distance from this point to the screen cannot be written in a simple algebraic expression,
but we are making only a relatively small error of < 0.5 cm if we assume that it is at a distance a = 10 cm
from the screen. With this approximation
tan θ
'
θ
'
w
= 2.5
a
680
¿From the figure we also see that
d
R
R
=
sin θ ' 0.93
'
1.08 cm
Inserting in (6.1) we find
r
B=
2 × 9.1 × 10−31 kg × 50, 000 V
1
= 0.070 T = 70 mT
1.6 × 10−19 C
0.0108 m
7. Serway 29-55
The Hall effect can be used to measure n, the number of conduction electrons per unit volume for an
unknown sample. The sample is 15 mm thick and when placed in a 1.8 T magnetic field produces a Hall
voltage of 0.122 µV while carrying a 12 A current.
a) Sketch the setup and what is happening.
b) What is the value of n for the sample?
SOLUTION:
a)figure to come In the sample shown below a current I is flowing from left to right which means that
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electrons flow to the left with average velocity vD . The magnetic field B is pointing vertically upward. The
electrons accumulate at the front edge of the conductor, positive charges are at the rear edge.
b)See page 852 in Serway. The Hall voltage is VH = vD Bd. The drift velocity is related to the current I, the
cross section td of the sample, and the number n of electrons per unit volume by I = nevD td. Combining
the two equations we find
n=
IB
12 A × 1.8 T
=
= 7.38 × 1028 m−3
e tVH
1.6 × 10−19 C × 0.015 m × 0.122 × 10−6 V
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