Physics 1302W.500, Complete Discussion Solutions
Scott Fallows
(Dated: January 28 – May 5, 2008)
I.
DECAYING SATELLITE ORBIT
A satellite in a circular Earth orbit is subject to a very small constant friction force f ,
due to the atmosphere. As it spirals inward, it slowly decreases its orbital radius. Find
the decrease in radius per revolution under the assumption that the orbit is approximately
circular with radius r. Find the changes in potential energy, kinetic energy, and total energy
per orbit. Does the satellite speed up or slow down as it spirals in?
Solution. For a circular orbit, the only radial (centripetal) force is gravity, so Fg = Fc
with Fg = GM m/r2 and Fc = mv 2 /r. This implies
mv 2
GM m
=
r2
r
r
=⇒
v=
GM
r
(1)
So as long as the satellite moves in a circular orbit, its velocity is inversely proportional to
the square root of the orbital radius – it speeds up as it spirals inward.
Angular momentum is not constant (~r × f~ = τ is an external torque), so we instead
consider energy. Using equation (1), the total energy is given by
E =T +U =
mv 2 GM m
GM m
−
=−
2
r
2r
(2)
The only nonconservative force acting on the system is friction, so the amount of energy
lost by the system in one revolution is equal to the amount of work done on the system by
friction over an approximate circle of radius ravg .
∆E = −Wf = −2πravg f
(3)
Energy is neither created nor destroyed, so we can write the conservation equation
−
GM m
GM m
− 2πravg f = −
2r1
2r2
(4)
This gives us a relation between the initial radius r1 and the radius after one revolution r2 .
We will approximate ravg two different ways. The first method is a bit more difficult, but it
2
allows us to input a starting radius and find the difference after one revolution. The second
method is simpler, but does not immediately allow for useful numerical results.
Method 1. Since the magnitude of the dissipated energy term is small compared to that
of the total energy and since ∆r is also small, we can approximate ravg with r1 and solve
for r2 in terms of r1 as follows:
r2 = r1
4πr12 f
+1
GM m
−1
(5)
2
Note that if r1 were at the surface of the Earth RE , we’d have the factor GM/RE
=g=
9.81 m/s2 in the denominator above, so the value at r1 should be of roughly the same order
of magnitude. For a force of friction f whose numerical value in N is much smaller than the
numerical value of the mass m in kg, we see that r2 = r1 ( + 1)−1 where 1. This can be
expanded in a Taylor series so that
r2 = r1
4πr12 f
+ ...
1−
GM m
(6)
In this form we can easily see that ∆r = r2 − r1 < 0.
The difference in kinetic energy ∆T = T2 − T1 = mv22 /2 − mv12 /2 so if we substitute in
equation (5) for r1 /r2 and cancel a factor of GM m/r1 we see that
GM m r1
GM m GM m
−
=
− 1 = 2πr1 f
∆T =
2r2
2r1
2r1
r2
The difference in potential energy ∆U = U2 − U1 is
GM m
GM m
GM m r1
∆U = −
− −
=−
− 1 = −4πr1 f
r2
r1
r1
r2
(7)
(8)
So, as expected, ∆E = ∆T + ∆U = −2πr1 f .
Method 2. If we instead make the approximation ravg =
√
r1 r2 (the geometric mean of
the two radii) we can write equation (4) as
GM m r2 − r1
GM m ∆r
−
=−
= 2πravg f
2
2
r1 r2
2
ravg
(9)
3
4πravg
f
(10)
GM m
This is actually a different statement, since our expression for ravg now contains r2 . Unlike
∆r = −
the first method, calculating ∆r now requires ravg , which itself requires r1 and r2 , so if we
had ravg we’d already have ∆r = r2 − r1 . To escape from this difficulty, we can further
approximate ravg ≈ r1 . Then ∆r in equation (10) becomes equivalent to taking the first two
terms in the Taylor expansion (6) and we recover our results from Method 1.
3
II.
CHARGED PENDULA IN EQUILIBRIUM
Two light, strong strings are attached to a horizontal support. At the other end of each
string hangs an object. One of the objects is known to have weight 2.000 N; the other is
unknown. A power supply is slowly turned on to give each object a charge, which causes
them to move away from each other. When operating level is reached, the two objects hang
at the same height, but at different angles. The string of the known mass makes an angle
of 10◦ with the vertical, and the other makes an angle of 20◦ . Find the unknown mass and
then estimate the net charge necessary for the observed deflection if the strings are 10 cm
long and the objects are taken to be spherical. List your assumptions.
Solution. We begin by defining quantities. Let d be the distance between the points
where the strings are attached to the support and let ` be the length of the string. Then
x1 = ` sin(θ1 ) is the horizontal distance the known mass is deflected from its initial position
and x2 = ` sin(θ2 ) is the horizontal deflection of the unknown mass. Fci is the Coulomb
force on the ith object.
Two methods will be demonstrated. The first is exact and relatively simple, but requires taking components of vectors. The second method makes a small approximation but
demonstrates how the use of earlier concepts can dramatically simplify new problems.
Method 1. The pendula are in equilibrium, so if we directly consider the forces on the
first we find
X
Fx = T1 sin θ1 − Fc1 = 0
(11)
X
Fy = T1 cos θ1 − m1 g = 0
(12)
Divide the first equation by the second to find
tan θ1 =
Fc1
m1 g
(13)
We then apply the same analysis to the second pendulum and find that tan θ2 = Fc2 /m2 g.
Since the Coulomb force of 1 on 2 is the same as the Coulomb force of 2 on 1, Fc1 = Fc2 . So
m2 = m1
tan θ1
= 0.09877 kg
tan θ2
(14)
We now assume the objects are spheres and turn to the Coulomb force on the first
explicitly, defining r as follows,
F =
1
q1 q2
1 q 1 q2
=
4π0 (x1 + x2 + d)2
4π0 r2
(15)
4
Using (13), we can now assume q1 q2 = q 2 and solve for q to find
q=
p
4π0 r2 m1 g tan θ1
(16)
We’re not given d, but if we take it to be 1 cm, we get q = 3.980 × 10−7 C. If we take d = 0,
then q = 3.359 × 10−7 C.
Method 2.
In the case of small perturbations a pendulum can be thought of as
a harmonic oscillator with spring constant k = mω 2 = mg/`.
This is entailed by
the standard pendulum approximation, that for small angles sin θ ≈ θ.
In our case,
sin 20◦ = sin 0.349 rad = 0.342 ≈ θ, so this approximation is fairly good. Hooke’s Law
tells us that the restoring force F~ = −k~x, so the magnitude of the spring force on mass i
is Fi = mi gxi /` = mi g sin θi . Again, the Coulomb forces between the two masses are of the
same magnitude. Since the only other force acting on each object on the horizontal direction
is the Hooke’s Law force, and since both pendula are in equilibrium, the restoring forces for
the two objects must be equal in magnitude.
m1 g sin θ1 = m2 g sin θ2
=⇒
m2 = m1
sin θ1
= 0.1035 kg
sin θ2
(17)
Compare this to (14). For small angles, sin θ ≈ tan θ, and our answers differ by only 4.6%.
As before, the Coulomb force should be equal to the Hooke’s Law force,
1 q1 q 2
= m1 g sin θ1
4π0 r2
We again assume that q1 q2 = q 2 and solve for q,
q=
p
4π0 r2 m1 g sin θ1
For d = 1 cm, q = 3.827 × 10−7 C. If d = 0, then q = 3.205 × 10−7 C. Again these numbers
differ from those found by Method 1 by about 4%, so our approximation that sin θ ≈ θ is a
good one.
III.
ION MOTION THROUGH A CHARGED RING
A He++ ion (often called an α particle) moves on axis toward a ring of charge Q = −8.0 µC
with radius R = 3.0 cm. It is on course to collide with a sample 2.5 mm from the ring, and
has velocity v0 = 200m/s when it passes through the center of the ring. Assume the radius
5
of the ring is much larger than the distance the ion will travel, R x. Will the ion reach
the sample? Note that e = 1.6 × 10−19 C, mHe = 6.7 × 10−27 kg.
Solution. This problem can be approached from two directions. Consideration of kinematics and dynamics leads directly toward the answer, but there’s a fair amount of algebra
and we have to make approximations. We will later consider energy, which uses concepts
that have not been formally introduced, but produces an exact answer very quickly.
Method 1. If we define the axis of motion to be the x-axis, then Newton’s equations of
motion describe the kinematics of the system:
1
x = x0 + v0 t + at2
2
v = v0 + at
Take as the initial condition the ion passing through the center of the ring and find x as
the particle comes to rest. Then x0 = 0 and v = 0. Newton’s second law gives us F = ma,
where the relevant force is Coulomb’s law for a ring of charge, experienced by an on-axis
charge q = 2e,
Fx =
x
qQ x
qQ
≈
4π0 (x2 + R2 )3/2
4π0 R3
where we have made the approximation that x2 + R2 ≈ R2 since x R. We now solve the
equations of motion and make the substitutions t = −v0 /a and then a = F/m into the x
equation to get
4π0 R3 mv02
x=
2qQ
1/2
= 561 nm
Method 2. Consider the potential energy U of the ion near a ring of charge. If we know
the on-axis electric potential V of a ring of charge, then
U = qV =
1
qQ
√
2
4π0 x + R2
~ = −∇V . It will be introduced in
This scalar potential V is related to the electric field by E
Chapter 24 of Fishbane.
We know the initial total energy Ei = Ti + Ui is the kinetic plus potential energy at the
center of the ring. This should be equal to the total energy when the particle has stopped
(when T = 0), so we have the conservation equation
Ti + Ui = Uf
=⇒
1
1 qQ
1
qQ
√
mHe v02 +
=
2
4π0 R
4π0 x2 + R2
6
Solve this for x and we eventually end up with
s
−2
2π0 mv02
1
2
x= R −
+
= 561 nm
qQ
R
This is the same answer as we found with Method 1. This indicates that the approximation
we made in Method 1 (that the radius of the ring is much greater than the distance the ion
will travel past the center) is very good. In both cases, the ion will not reach the sample.
IV.
BONUS PROBLEM: FISHBANE 21.55
A total charge Q = 0.75 µC is distributed uniformly over a thin, semicircular wire of
radius R = 5.0 cm. What is the force on a charge of q = 0.30 µC located at the center of
the circle?
Solution. Coulomb’s Law is defined for point charges, so to find the force for a continuous
distribution of charge, we have to break the distribution into pointlike infinitesimal pieces
of charge dQ and integrate. Additionally, if we define our coordinates so the semicircle is
located in the +x half of the x-y plane with the point charge at the origin, we see that the
symmetry of the system implies that all the y components of force will cancel. We are left
with the x components, which we get by observing that Fx = F cos θ. So the total force is
Z
Z
q cos θ
1
dQ
Fx = dFx =
4π0
R2
The problem is now reduced to finding a way to express dQ such that we know how to
integrate over it.
For a linear charge distribution, dQ = λdl, where λ is the linear charge density and dl is
the infinitesimal length element. For a line charge of total length πR and total charge Q,
λ = Q/πR.
Now consider the element of length along the semicircle. The definition of angle is arc
length divided by radius, so dl = Rdθ. Finally, we know how to integrate over dθ, from
−π/2 to π/2.
qQ
Fx = 2
4π 0 R2
Z
π/2
cos θdθ =
−π/2
qQ
2π 2 0 R2
Clearly the charge will experience a force in the direction opposite to the direction of the
ring. This can be seen in that r̂ will be pointing in the −x direction, so this is the direction
of the total force.
7
V.
ELECTRIC FIELD OF OPPOSITELY CHARGED SEMICIRCLES
A charged ring of conductor is divided into two half circles separated by a thin insulator
so that half of the circle is charged positively and half is charged negatively. An electron
beam goes through the center of the circle. Find the electric field in the center of the circle
as a function of the amount of positive charge on one half circle, the amount of negative
charge on the other, and the radius of the circle.
Solution. This is essentially the same problem as Fishbane 21.55, except we now want
the field rather than the force on a specific charge, and there are two half circles. We’ll
consider the field due to one half circle and then the other. Start with the positive half. The
electric field is
Z
Ex =
1
dEx =
4π0
Z
cos θ
dQ1
R2
Recall that dQ = λdl = (Q/πR)dl = (Q/π)dθ. So we want to integrate
Z 3π/2
Q1
Q1
cos θdθ = 2
Ex = 2
2
4π 0 R π/2
2π 0 R2
Note that there is some arbitrariness in sign; that is, the direction we integrate around the
half circle. The easiest way to decide the direction is to do is a reality check: if Q1 is positive
the field points away from it, and vice versa.
Now we see that a similar integration gives us the contribution from the other half-circle,
Z π/2
Q2
Q2
Ex = 2
cos
θdθ
=
4π 0 R2 −π/2
2π 2 0 R2
Thus the total electric field at the center of the circle is
Ex,total =
Q1 + Q2
2π 2 0 R2
The direction is toward the negatively charged half-circle.
VI.
TRITIUM IONS IN A PLASMA-FILLED CYLINDER
A device confines a hot gas of positively charged ions, called plasma, in a very long
cylinder with a radius R = 2.0 cm. The charge density of the plasma in the cylinder is
ρ = 6.0 x 10−5 C/m3 . Positively charged tritium ions are to be injected into the plasma
perpendicular to the axis of the cylinder in a direction toward the center of the cylinder.
8
Find the speed that a tritium ion should have when it enters the plasma cylinder so that its
velocity is zero when it reaches the axis of the cylinder. Tritium is an isotope of Hydrogen
with one proton and two neutrons. Note that the charge of a proton and the mass of tritium
are 1.6 × 10−19 C and 5.0 × 10−27 kg.
Solution. We use Gauss’s Law to find the electric field inside the cylinder, noting that
the electric field is parallel to the top and bottom of our cylindrical Gaussian surface S of
radius r < R and arbitrary length l, so they contribute no flux.
I
~ · dA
~ = Qenc
E
0
S
Symmetry implies that E is constant around the cylinder at r, so it can be pulled out of the
integral, which can then be evaluated and combined with Qenc = ρV = ρπr2 l to yield
E · 2πrl =
ρπr2 l
0
=⇒
E=
ρr
20
The strategy will now be to equate the kinetic energy lost by the tritium ion to its increase
in electrical potential energy. We can write U = qV where
Z r
2
~ · dr~0 = − ρr
E
V =−
40
0
Conservation of energy now tells us that
mv 2 qρR2
0 = ∆T + ∆U = − 0 +
2
40
VII.
=⇒
v0 =
qρR2
2m0
1/2
= 456 m/s
ELECTRIC FLUX THROUGH A CUBE
A given region has an electric field that is the sum of two parts: a field due to a point
charge q = 5 ×10−8 C at the origin plus a uniform electric field of magnitude 3000 N/C
directed in the −x direction. Calculate the flux through each face of a cube with sides of
length 20 cm and centered on the origin. The faces are parallel to the x, y, z axes.
Solution. The uniform field is parallel to the unit normal vector of the surface at x = −10
cm, antiparallel to the one at x = +10 cm, and perpendicular to all the rest, so the flux
due to the uniform field is just Φunif = 3000 N/C × (0.2 m)2 = 120 Nm2 /C on one face and
similarly -120 Nm2 /C on the other. We now turn to the flux from the point charge.
Method 1. The electric field from the point charge is
1 q
1
q
E=
=
4π0 r2
4π0 x2 + y 2 + z 2
9
Consider the surface at z = 10 cm. The flux through this surface will be
Z
Z 10 Z 10
cos θ
q
~
Φchg = (n̂ · E)dA =
dxdy
2
4π0 −10 −10 x + y 2 + z 2
S
Where θ is the angle from the z axis to point of integration. To do this integral we need to
write cos θ = z/r to find
q
ΦS =
4π0
Z
10
−10
Z
10
−10
(x2
z
q
× 2.0944
dxdy =
2
3/2
+ +z )
4π0
y2
Due to symmetry, all six sides contribute the same flux. Notice that 2.0944 × 6 = 4π. So
the total flux from the charge is q/(4π0 ) × 4π = q/0 . This suggests another method we
could have used to simplify the problem.
Method 2. Use Gauss’s Law, considering the cube as a closed surface that encloses the
charge q. Then Gauss’s Law says that the flux of the charge will be
I
~ · dA
~= q
E
0
cube
This immediately gives us the same answer for the flux of the charge through the cube that
we found using brute force in Method 1, with one sixth of the flux passing through each face
of the cube. The positive and negative contributions from the uniform field cancel when
summed and the final answer is
Φtotal = Φunif + Φchg = (0 + 5647) Nm2 /C = 5647 Nm2 /C
VIII.
ACCELERATION TOWARD A LINE CHARGE
A particle enters an electrostatic trap and is exposed to ultraviolet radiation that knocks
off electrons so that it has a charge of 3.0 × 10−8 C. This particle is then moving with speed
v0 = 900 m/s and is 15 cm from a very long negatively charged wire with a linear charge
density λ = −8.0 × 10−6 C/m. The detector for the particle is located 7.0 cm from the wire.
The particle has a mass of 6.0 × 10−9 kg. Find the speed of the particle just before it hits
the detector.
Solution. This problem can be best dealt with using conservation of energy. To find the
potential energy of the particle near the line charge, we first use Gauss’s Law to find the
electric field at a distance r from the axis,
E(r) =
λ
2π0 r
10
Integrate the field to find the potential, taking the zero potential reference at r = a,
Z r
r
λ
E(r0 )dr0 = −
V (r) = −
ln
2π0
a
a
So if a is the distance from the line to the detector, and if r is the distance from the line to
the charge’s initial position, the above expression is the potential difference between these
two points ∆V and the change in potential energy is given by
r
qλ
ln
∆U = q∆V = −
2π0
a
Conservation of energy tells us that ∆U + ∆T = 0, so
∆T =
m 2
vf − v02 = −∆U
2
Solving for vf , we find that
r
vf =
v02 −
r
qλ
ln
= 1381 m/s
mπ0
a
Note that the negative sign of λ makes the second term under the radical positive, resulting
in vf > v0 , as expected.
IX.
COLLISION OF ALPHA PARTICLES WITH LEAD NUCLEI
You shoot α particles from a Van de Graaf accelerator at a sheet of lead. The α particle
is the nucleus of a helium atom and is made of 2 protons and 2 neutrons. The lead nucleus
is made of 82 protons and 125 neutrons. The mass of the neutron is almost the same as
the mass of a proton. The α particle should come into contact with a lead nucleus. Assume
that both the α particle and the lead nucleus have the shape of a sphere. The α particle
has a radius of 1.0 × 10−13 cm and the lead nucleus has a radius 4 times larger. What is
the minimum speed of such an α particle if the lead nucleus is fixed at rest? What is the
potential difference between the two ends of the Van de Graaf accelerator if the α particle
starts from rest at one end? The mass of a proton mp ≈ 1.67 × 10−27 kg.
Solution. Initially, assume the electric potential energy between the α particle and the
82
Pb nucleus is zero. For the minimum approach speed, all initial kinetic energy will be
converted into potential energy upon contact. Since the nuclei are assumed to be spherical,
we can treat all the charge as being located at the center of each. Then contact will be made
11
when the centers of the nuclei are 5rα apart. The charges of the α particle and the
82
Pb
nucleus are 2e and 82e, respectively. So the potential energy when the two make contact is
U=
Solve for v0 to find
s
v0 =
(2e)(82e)
mα v02
=
(4π0 )5rα
2
328e2
= 4.75 × 107 m/s
(4π0 )5mα rα
The kinetic energy of a charge accelerated through a voltage is T = qV so
T
mv02
V =
=
= 2.36 × 107 V
2e
4e
Note that real Van de Graaf generators can reach approximately 5 MV.
X.
CAPACITOR-POWERED ELECTRIC CAR
You want to investigate the practicality of a plug-in hybrid car that can go some distance,
say 30 miles, on stored electrical energy before having to switch on a liquid fuel powered
engine. You know that it takes about 0.2 kW-hr/mile to power the car. You want to know
what the parameters would be if the electrical power source is a bank of capacitors arranged
in parallel. Suppose that you are told that a realistic voltage for a capacitor is 400V. What
is the required total capacitance of the bank of capacitors?
Since it is not at all unusual for electric cars to use a very large number of individual
3.7 V LiIon batteries (hundreds or thousands), suppose that you can use 500 individual
capacitors. Think up some parameters (total plate area, separation, dielectric constant)
for an individual capacitor in the bank that do not seem too crazy. Could this bank alone
maintain a constant 400 V as it discharges, or would some additional circuitry be needed?
Solution. The energy needed to travel 30 miles is given by
(0.2 kW-hr/mile) ∗ (30 miles) = 6 kW × 3600 s = 21.6 × 106 J.
Since the energy stored in a capacitor is U = CV 2 /2, the total capacitance required to store
the necessary energy is
C=
2U
= 270 F
V2
12
Note that this is the capacitance needed to deliver the necessary amount of energy when
initially charged to the required voltage, but the voltage will decrease as the capacitor is
discharged, to below the value we’ve required. Not a good sign for this concept working out.
Let’s make some generous estimates on the capacitance C = A/d. Take the dielectric
constant /0 to be 3, the plate area to be 1 m2 , and the separation to be 10−5 m. Then the
capacitance is
C=
A
= 2.66 × 10−6 F
d
Since the capacitors are connected in parallel, the total capacitance would be CT = 500C =
1.33 × 10−3 F. This is nowhere near 270 F, so the car won’t be able to maintain 400 V very
long, certainly not the whole trip.
XI.
DETERMINATION OF A WIRE’S LENGTH
When 12 V is applied across a wire of unknown length, a current of 0.06 A flows. When
1.5 V is applied aross a 100-foot wire of the same cross-sectional area and conductivity as
the first, 0.10 A flows. Find the unknown length.
Solution. We can write the resistance of a conducting wire in terms of its length, crosssectional area, and conductivity as R = L/σA. The wires should be roughly ohmic, so we
can write V = IR. Label the mystery wire 1 and the test wire 2, then take the ratio of L1
to L2 . The constants σA drop out and we are left with
R1
V1 I2
L1
=
=
L2
R2
V2 I1
Solve for L1 and plug in the given values to find L1 = 1333 ft = 406.4 m.
XII.
CURRENT DRAWN BY APPLIANCES
A 1000-watt waffle iron and a 600-watt coffee maker are plugged into 110 V kitchen outlets. If you plug in a 700-watt blender, will you overload the 20 A circuit breaker? Note that
the circuit topology of the outlets can be determined based on well-known characteristics:
a connected device will still be powered even if one of the outlets is not being used.
Solution. For moving charges to be able to complete the circuit without devices in all
three outlets, the outlets must be wired in parallel. This means that each device will have
13
110 V applied across its terminals. Each appliance will draw the amount of current down
its branch of the circuit that it needs to generate its stated power consumption.
Imagine the kitchen outlets supply 110 V DC (it’s actually AC, which will be covered
later; the current can be run through a rectifier to provide the DC we have in this problem).
Then the current drawn by each device can be found by the relation P = IV . Plugging in
the stated power requirements and the constant voltage applied across each of the devices,
Iwaffle =
1000 W
Pwaffle
=
= 9.09 A
V
110 V
600 W
Pcoffee
=
= 5.45 A
V
110 V
Pblender
700 W
=
=
= 6.36 A
V
110 V
Icoffee =
Iblender
All these currents are being drawn from the same source, so before branching into the three
parallel segments of the circuit, the wire is carrying a total current
IT =
blender
X
Ii
i=waffle
So IT = 20.9 A, which is enough to overload the 20 A circuit breaker.
XIII.
HEAT GENERATION IN A DC CIRCUIT
A second 12 V battery is added to a sensitive low-temperature device. Determine the
heat generated by the 20 Ω resistor in the modified circuit below.
Solution. Label the resistors: R1 = 100 Ω, R2 = 20 Ω, R3 = 40 Ω, R4 = 200 Ω. To find
the heat generated by R2 , we need to find the current through it and then use it to find the
rate of energy being dissipated in the resistor, P = I 2 R. This is easier if we first combine
the parallel resistors into R = R2 R3 /(R2 + R3 ), find the current through R, and then use
the current divider relation to find the fraction of this current going through R2 .
The equivalent circuit now has two nodes (one independent equation from the Kirchhoff
Junction Rule) and two independent loops (two independent equations from the Kirchhoff
Loop Rule). Label the current going clockwise in the left branch I1 , the current going down
in the middle branch I, and the current going clockwise in the right branch I3 . Then the
junction rule gives us
I1 = I + I3
(18)
14
Now use the loop rule on the left (clockwise) and right (counter-clockwise) sub-loops to find
V − IR − I1 R1 = 0
(19)
V + I3 R4 − IR = 0
(20)
Substitute (18) into (19), then solve (20) for I3 and substitute it back into the modified (19).
This gives
IR − V
V − IR − I +
R4
R1 = 0
which we multiply by R4 and solve for I to get
I=
V (R1 + R4 )
= 150 mA
RR1 + RR4 + R1 R4
Now we need to find the amount of current going through R2 . Expand the middle branch
back to two resistors in parallel. Then the current divider relation gives
I2 =
R3
I = 100 mA
R2 + R3
The dissipated power is P = I22 R2 = 0.2 W.
XIV.
POWER RATING FOR RESISTORS IN A DC CIRCUIT
The resistors in the following circuit are rated at 0.5 W, meaning they burn up if more
than 0.5 W of power passes through them. Will the 100 Ω resistor burn up?
Solution. First redraw the circuit so that the upper-left triangular loop is a rectangular
loop on the left and the lower-right triangular loop is a rectangular loop on the right; this
makes the problem easier to visualize. Now the problem is essentially the same as the
previous, with slightly different circuit elements present. Again, we need to find the current
through a resistor to determine the power it dissipates.
Label the components: R1 = 100 Ω, R2 = 133 Ω, R3 = 200 Ω, V1 = 9 V, V2 = 6 V. Call
I1 the clockwise current in the left branch, I2 the downward current in the middle branch,
and I3 the clockwise current in the right branch. Consider the node to the right of R1 . The
Kirchhoff Junction Rule gives
I2 = I1 − I3
(21)
15
The Kirchhoff Loop Rule going clockwise in the two smaller independent loops gives
V1 − I1 R1 − I2 R2 = 0
(22)
V2 − I3 R3 + I2 R2 = 0
(23)
Add (22) to (23) and solve for I3 to get
I3 =
V1 + V2 − I1 R1
R3
(24)
Substitute (21) into (22) and then substitute (24) into the modified (22). This gives
V1 + V2 − I1 R1
R2 = 0
V1 − I1 R1 − I1 −
R3
which we multiply by R3 and solve for I1 to get
I1 =
V1 (R2 + R3 ) + V2 R2
= 43.3 mA
R1 R2 + R1 R3 + R2 R3
The dissipated power is P1 = I12 R1 = 0.40 W, so the 0.5 W resistor R1 will not burn up.
XV.
DEFLECTION BY THE EARTH’S MAGNETIC FIELD IN A CRT
Electrons are boiled off of a cathode at the back of a CRT and are accelerated through
a potential of 20,000 V toward the screen. Pixel separation on the screen is 1/100th of an
inch. Does such a CRT need to be shielded from Earth’s magnetic field? Note the following
constants: qe = -1.602 ×10−19 C, me = 9.109 ×10−31 kg. The strength of Earth’s magnetic
field is about 0.5 gauss at the surface, or 50 µT.
Solution. The electrons (“cathode rays”) each gain a kinetic energy of mv 2 /2 = qV and
p
so they have velocity v = 2qV /m = 8.387 × 107 m/s. If we assume the length of the tube
is about half a meter, it takes t = d/v = 5.96 ns for an electron to travel from one end to
the other.
We now have everything we need to determine the deflection due to the Earth’s magnetic
field using the Lorentz Force Law:
~ = qvB sin θ
F~ = q~v × B
~ We determined previously that the effect of gravity
where θ is the angle between ~v and B.
is negligible over so short a time (∆y = gt2 /2 = 1.7 × 10−16 m here) so we will consider only
16
the magnetic force on the particle. Orient the tube so that the magnetic field is as directly
perpendicular to the particle trajectory as possible, so sin θ = 1 and the Lorentz force is
~ call it ẑ. Then
maximum. The direction of the deflection is perpendicular to both ~v and B;
∆z = vz,0 t +
Fz t2
qvBt2
az t2
=0+
=
= 1.31 cm
2
2m
2m
So depending on the orientation of the tube, the electrons may be deflected by an amount
on the order of a centimeter, or about 50 pixel widths on the TV in question. Because of
this (and other sources of magnetic fields, particularly internal circuitry), all CRTs used in
televisions have internal magnetic shielding.
XVI.
MAGNETIC FIELD FROM A PART-CIRCULAR CURRENT LOOP
A current runs counter-clockwise along a circular wire segment of radius b beginning at
point D and centered on point P , forming an arc of 120◦ . The wire then bends and current
flows radially outward for a length a. Current then flows clockwise along a 120◦ arc of radius
a+b centered at P , returning to the same angular position as D, then moving radially inward
to meet up with D. Find the magnetic field produced at P .
Solution. For this problem we will make use of the Biot-Savart Law,
~ =
dB
µ0 Id~` × ~r
4π r3
(25)
The line segments AB and CD are parallel to the radius vector from point P, so for these
segments, d~` ×~r = 0. On the circular portions of the loop, |d~` ×~r| = rd` = r(rdθ). Consider
the inner loop of radius b. For convenience, define the angles relative to the line P C.
Z 2π/3
Z 2π/3
µ
I
µ0 I
0
~ inner =
Binner =
dB
d` =
2
4πb 0
6b
0
By the right hand rule, this field is pointing out of the page. Now consider the outer loop,
Z 2π/3
Z 0
−µ0 I
−µ0 I
~
Bouter =
dBouter =
d` =
2
4π(a + b) 0
6(a + b)
2π/3
where the negative sign means the field points into the page, by the angular convention
we’ve adopted for this problem. Use a common denominator to add the fields and find that
B = Binner + Bouter =
µ0 Ia
6b(a + b)
17
The sign indicates that field points out of the page. This is as we should have expected,
since the inner arc has a smaller radius and hence a larger contribution to the field.
We could have done this problem more carefully by checking the sign of d~` × ~r for each
arc and making sure we’re integrating in the correct direction, while taking account of the
direction of current, but it’s much easier to just look at magnitudes and deduce the final
direction from what we know the result has to look like.
XVII.
MAGNETIC FIELD FROM A COAXIAL CABLE
A coaxial cable consists of an inner wire surrounded by an insulator, followed by a conducting shell. The wire and shell both carry the same current, but in the opposite direction.
Find the magnetic field at a radius larger than that of the shell and at a radius smaller than
that of the shell. Then take the shell to have non-negligible thickness and find the magnetic
field between inner radius R1 and outer radius R2 of the shell.
Solution. For this problem we will make use of Ampère’s Law,
I
~ · d~s = µ0 Ienclosed
B
(26)
Outside the conducting shell, the total enclosed current Ienc = 0, so B = 0. In the region
between the central wire and the outer shell, draw an Ampèrian loop of radius r. Then
B(2πr) = µ0 I
=⇒
B=
µ0 I
2πr
Now we wish to consider the magnetic field inside the outer shell, between inner radius
R1 and outer radius R2 . The problem is trivial if we assume that all the current moves on
the outer surface of the conductor, and the previous answer applies.
If we make it interesting by assuming the current is uniformly distributed throughout the
shell, then our loop of radius r such that R1 < r < R2 will enclose a total current of I − f I
where f accounts for the fact that only part of the current in the outer shell is enclosed,
and is the ratio of the current-carrying area of a circle of outer radius r to that of a circle
of outer radius R2 (both current-carrying areas have inner radius R1 ). That is,
f=
πr2 − πR12
πR22 − πR12
Then the amount of current enclosed is
r2 − R12
=I 1− 2
R2 − R12
Ienc
18
Note that this gives us the correct boundary conditions: at r = R1 , Ienc = I and at r = R2 ,
Ienc = 0.
Finally, we plug back into (26) to find that
B=
XVIII.
µ0 Ienc
=
2πr
µ0 I 1 −
r2 −R12
R22 −R12
2πr
BAR SLIDING ON INCLINED RAILS IN A MAGNETIC FIELD
A conducting bar slides on two parallel conducting rails that run down a ramp. At the
bottom of the ramp, the rails are connected. The bar slides down through a uniform vertical
magnetic field that causes the bar to slide down the ramp at constant velocity even when
friction between the bar and rails is negligible. Find this constant velocity.
Solution. This problem has some tricky geometry but is conceptually identical to Fishbane 30.58(a). First, let’s define the necessary parameters:
• v, the constant speed of the bar
• m, the mass of the bar
• B, the strength of the magnetic field
• θ, the angle of the ramp from the horizontal
• L, the length of the bar
• R, the resistance of the bar
~ the normal area vector of the loop formed by the bar and rails
• A,
• s, the distance along the ramp from the base to the bar
Call the direction along the incline ŝ. We’re given that the velocity is constant, so by
P
Newton’s second law
Fs = mas = mg sin θ − FB,s = 0. So the component of the force of
gravity in the ŝ direction balances the ŝ component of the magnetic force due to the change
in magnetic flux through the loop as given by Faraday’s Law,
E =−
dΦB
,
dt
(27)
19
the Lorentz force,
F~B =
Z
~ =I
dq(~vq × B)
Z
~
d~` × B,
(28)
and Ohm’s Law, E = IR.
Note that (28) references ~vq , the velocity at which the charges move; this is not the droid
v we’re looking for. Since rate of change of magnetic flux through the loop dΦ/dt depends
on the changing area of the loop, it will depend on the rate at which the bar is moving, that
is, decreasing the area. This will give us a v dependence that we can solve for.
Combining (27), (28), and Ohm’s Law,
Z
Z
dΦ
1
E
~ =−
~
d~` × B
d~` × B
F~B =
R
R dt
We’re considering the force on the bar, so the integrand is a vector pointing into the ramp
in the horizontal direction. So the component of the force along the +ŝ direction is
1 dΦ
LB cos θ
FB,s = FB cos θ = −
R dt
(29)
We now turn our attention to the time rate of change of the magnetic flux through the
~ parallel to the magnetic field, then Φ = B
~ ·A
~ = BAk . If
loop. If Ak is the component of A
the magnitude of the area A = Ls, then Ak = Ls cos θ. So
dAk
ds
dΦ
=B
= BL cos θ = −BLv cos θ
dt
dt
dt
If we insert this result into (29), we find that
FB,s =
1
(BLv cos θ) LB cos θ
R
Set this equal to mg sin θ and solve for v:
v=
XIX.
mgR sin θ
(BL cos θ)2
MAGNETIC FIELD IN AN IRON-FILLED TOROID
A torus with a central radius of 25 cm and a tube radius of 2.0 cm is filled with iron
of susceptibility 2800. There are 1200 turns around the torus. How much current must
flow in the winding coil to produce a field of 1.5 T inside the torus? You can take B to be
approximately constant and equal to the value on the axis of the torus.
20
Solution. Using Ampère’s Law, create a loop around N windings with length L. Then
I
B~0 · d~` = B0 L = µ0 Ienc = µ0 N I
Assume the density of the windings is constant, so define n = N/L, so B0 = µ0 N I/L = µ0 nI.
The magnetic field inside the torus is stronger due to the iron:
B = (1 + χm )B0 = (1 + χm )µ0 nI = µnI
(30)
Solve for the current and expand n to the full circumference and total number of turns to
find
I=
B
B(2πR)
B(2πR)
=
=
= 558 mA
µn
µN
(1 + χm )µ0 N
Note that (30) is a general equation for solenoid-type objects filled with material of arbitrary
permeability. Though it doesn’t take long to derive, it’s worth memorizing.
XX.
SIMPLE LC CIRCUIT
Consider a circuit consisting of a capacitor (C = 20 nF) and an inductor (L = 0.02 mH).
An initial charge (Q0 = 20 nC) is placed on the capacitor, with no current in the circuit. A
current then starts to flow. Find the maximum current in the circuit, the maximum energy
stored in the inductor, and the maximum energy stored in the capacitor.
Solution. An application of the loop rule to the circuit yields the homogeneous, linear,
second-order differential equation
0=
Q
dI
Q
d2 Q
+L = +L 2
C
dt
C
dt
=⇒
d2 Q
= −ω 2 Q
dt2
√
where ω = 1/ LC, which has solutions
Q(t) = Q1 sin(ωt) + Q2 cos(ωt)
Our initial condition Q(t = 0) = Q0 tells us that Q1 = 0 and Q2 = Q0 . So
Q(t) = Q0 cos(ωt)
(31)
The current is then
I(t) =
dQ(t)
= −ωQ0 sin(ωt)
dt
(32)
21
Since −1 < sin(ωt) < 1, the maximum value of the current is Imax = ωQ0 =
√Q0
LC
= 32 mA.
We now consider the energy stored in the capacitor,
UC =
CV 2
Q2
Q2
=
= 0 cos2 (ωt)
2
2C
2C
Again, the cosine is bounded by ±1, so UC,max = Q20 /2C = 10−8 J.
Similarly, the energy stored in the inductor is maximal when the current is maximal, so
UL,max =
2
1
Q2
LImax
= Lω 2 Q20 = 0 = 10−8 J
2
2
2C
This just tells us what we already knew: all the energy initially stored on the capacitor has
to be stored in the inductor when the capacitor’s charge is zero, since there is no dissipative
element (like a resistor) in this circuit, and energy is conserved. Eq. (31) and (32) tell
us that the charge on the capacitor is zero when the current is maximal, so everything is
consistent.
XXI.
HARMONICALLY DRIVEN LC CIRCUIT
Consider a series ac circuit consisting of a voltage supply oscillating at 600 Hz and whose
emf has magnitude V0 = 4 V, two capacitors in series with C1 = 4 µF and C2 = 9 µF, and
an inductor with L = 70 µH. Find the maximum current and the resonance frequency.
Solution. First combine the capacitors into an equivalent capacitance of magnitude
C = C1 C2 /(C1 + C2 ). Application of the Kirchhoff Loop Rule yields
V0 sin ωt − L
dI
Q
− =0
dt
C
(33)
This is much more difficult to solve explicitly than the equation from last week’s problem,
but there is a way to dramatically simplify the process. We know that eventually, the
oscillatory behavior of the circuit will be entirely governed by the driving voltage, and so
the final solution for the current in the circuit (or the charge on a capacitor) will have the
same general form as the driving voltage,
Q(t) = Qmax sin ωt
(34)
If we plug this trial solution (34) back into (33), realizing that I = dQ/dt, we get
V0 sin ωt − L(−ω 2 Qmax sin ωt) −
Qmax
sin ωt
C
22
Divide out the sin ωt factor and solve for Qmax ,
Qmax =
V0
1/C − Lω 2
(35)
Then we can find
I(t) =
dQ(t)
= ωQmax cos ωt
dt
So Imax = ωQmax = 41.9 mA.
The resonance (natural) frequency in a simple LC circuit is independent of the driving
voltage and so is easy to remember, but given our previous solution, it’s straightforward to
find. Resonance is characterized by a sharp peak in the response (current, charge, etc.) of
a circuit at a finite frequency. To make the expression for Qmax blow up, we want to make
the denominator 0. Solving for the unique positive ω that does this results in,
1
= 71814 s−1
ω0 = √
LC
Note that we can now rewrite (35) as follows:
Qmax =
V0
− ω2)
L(ω02
This clearly shows the expected resonance behavior: as ω gets arbitrarily close to ω0 , the
charge on the capacitor gets arbitrarily large. This puzzling situation is resolved if you add
a small resistor (as there must be in any real situation), which adds another term to the
denominator.