http://iml.umkc.edu/physics/wrobel/phy250/homework.htm
Homework 9
chapter 29: 23, 29, 35, 61
Problem 29.23
The picture in a television uses magnetic deflection coils rather than electric
deflection plates. Suppose an electron beam is accelerated through a 50 kV
potential difference and then through a region of uniform magnetic field 1 cm
wide. The screen is located 10 cm from the center of the coils and is 50 cm wide.
When the field is turned off, the electron beam hits the center of the screen. What
field magnitude is necessary to deflect the beam to the side of the screen?
L
r
α
y
V
D
The electrons leave the gun with a speed dependent on the
potential difference. From the work-energy theorem the speed of
the electrons is
v=
2 Wel
2eV
=
m
m
where m is the electron mass, e the elementary charge and V the
accelerating potential difference.
When the electrons enter the magnetic field, they move
along a circular path due to the magnetic interaction. The
radius of the path can be determined from Newton’s second law
– the centripetal acceleration is proportional to the magnetic
force
r=
mv
eB sin 90°
2
The radius of the circular path (and the width of the magnetic
field) affects the deflections angle
sin α =
D
r
That angle also determines where the electron beam hits the
screen
y ≈ L ⋅ tan α
Hence
B=
mv
=
er e ⋅
m⋅
2eV
m
D
sin arctan
=
2meV
y
⋅ sin arctan
L
eD
y
L
For the beam to reach a side of the screen, the magnetic field
strength must be
Bmax =
(
)(
) ⋅ sin arctan 25cm = 70.4mT
10cm
C ) ⋅ 0.01m
2 ⋅ 9.1 ⋅ 10 − 31 kg ⋅ 50 ⋅ 103 V
(1.6 ⋅10
−19
3
Problem 29.29
A rod of mass 0.720 kg and radius 6 cm rests on two parallel rails that are 12 cm
apart and 45 cm long. The rod carries a current of 48 A (in the direction shown in
the figure) and rolls along the rails without slipping. A uniform magnetic field of
magnitude 0.240 T is directed perpendicular the rod and the rails. If it starts from
rest, what is the speed of the road as it leaves the rails?
B
d
I
F
L
The simplest solution is based on the work energy theorem.
Three objects interact with the rod – the earth (gravity), the rails
(normal and static friction) and the (source of) magnetic field
(magnetic). Assuming that the rails are horizontal, only the
magnetic interaction results in non-zero total work
Wnet = WB = ∫ FB ⋅ ds = I ⋅ d ⋅ B ⋅ sin 90° ⋅ L ⋅ cos 0° = IBLd
path
This work must be equal to the change in the total kinetic energy
of the rod.
K tot = K rot
2
1 ⎛1
Mv 2 3Mv 2
2⎞v
+ K trans. = ⋅ ⎜ Mr ⎟ 2 +
=
2 ⎝2
2
4
⎠r
Therefore
v=
4 Wnet
4IBLd
4 ⋅ 48A ⋅ 0.24T ⋅ 0.45m ⋅ 0.12m
m
=
=
= 1.07
3M
3M
3 ⋅ 0.72kg
s
(Note that the radius of the rod has no effect on the speed.)
4
Problem 29.35
A rectangular loop consists of N = 100 closely wrapped turns and has dimensions
a = 0.40 m by b = 0.30 m. The loop is hinged along the y axis, and the plane of the
coil makes an angle of θ = 30° with the x axis (Fig. P29.25). What is the magnitude
of the torque exerted on the loop by a uniform magnetic field of B = 0.8 T directed
along the x axis when the current has a value of I = 1.2 A in the direction shown?
What is the expected direction of rotation of the loop?
y
The magnetic torque exerted on
an object with magnetic moment μ is
related to the magnetic field vector
(definition of the magnetic moment):
(1)
τ = μ×B
I=1.2A
μ
a =0.4m
B
τ
x
α =30
b =0.3m
z
From the text of the problem we
know that the magnetic field is in the
x-direction:
B = B, 0, 0
(2)
We have to find the magnetic moment of the loop. The
direction of the loop's magnetic moment is perpendicular to the
surface of the loop following the right-hand rule. I marked the
direction in the figure. Its magnitude depends on the number of
turns, the electric current in the wire and the area of the loop.
μ = nIab
Consistent with the choice of the coordinate system, the
magnetic moment is therefore:
μ = nIabsin α , 0, − nIabcos α
(3)
Using (2) and (3) in (1) we find the magnetic torque:
5
τ = μ yBz − μ z By , μ z Bx − μ x Bz , μ x By − μ yBx = 0, − nIBabcos α , 0 =
= 0, − 100 ⋅1. 2 A ⋅ 0.8T ⋅ 0. 4 m ⋅ 0.3m ⋅ cos 30o , 0 ≈ 0,10, 0 Nm
Recalling Newton's second law, we can relate the net
torque with the rate of change in the angular momentum.
dL
= τ net
dt
The hinges and the gravity cause a horizontal torque, which
cannot affect the vertical component of the angular momentum.
The hinges constrain the loop to rotation about the y-axis. For
a rectangular loop, the direction of angular momentum L
coincides with the direction of angular velocity ω. Considering
the y-components we obtain:
dω
Ia
=τ
dt
where Ia is the moment of inertia of the loop about the axis of
rotation. If initially the loop did not rotate, the magnetic
moment will cause it to rotate with angular velocity down.
6
Problem 29.61
A heart surgeon monitors the flow rate of blood through an artery using an
electromagnetic flowmeter. Electrodes A and B make contact with the outer
surface of the blood vessel, which has interior diameter 3 mm. (a) For a
magnetic field magnitude of 0.04 T, an emf of 160 μV appears between the
electrodes. Calculate the speed of the blood. (b) Verify that electrode A has
higher potential, as shown. Does the sign of the emf depend on whether the
mobile ions in the blood are predominantly positively or negatively charged?
Explain.
a) The ions move along the
vessel while the magnetic
force is transverse to the
vessel. The redistritbution of
ions results in a transverse
electric field resulting in a
force on the ions in the
opposite direction
Artery
+
A
F
B
N
v
-
B
to voltmeter
S
Blood
flow
qE = − qv × B
or in terms of magnitudes
qE = qvB sin 90°
Hence the speed of the ions must be
v=
160μV
m
E V
=
=
= 1.33
s
B Bd 0.04T ⋅ 3mm
b) Consistent with the right hand rule, the direction of the
magnetic force exerted on the ions with a positive charge is
upward while the force exerted on the negatively charged ions is
downward. In both cases, the polarity of the electric field is
consistent with the indicated potential of the electrodes. Both
types of ions contribute to the electromotive force.
7