Homework Solutions
Chapter 22
(1) “Motional EMF:” E = v B L . This is true for cases in which the velocity v, the
magnetic field vector B, and the length L are all mutually perpendicular (like the X, Y,
and Z axes in a standard coordinate system. In this case, we are told that the airplane
flies “horizontally” (so its velocity vector is horizontal); we must assume that it also flies
with its wings level, which would ordinarily be the case, so that the length of its wings is
also horizontal – but, of course, perpendicular to the velocity vector. We are told that the
magnetic field vector we’re given is the vertical component of the earth’s B. So, the
condition of mutual perpendicularity is satisfied, and we can substitute and do arithmetic:
E = v B L = (220 m/s) (5.0 x 10-6 T) (59 m) = 6.5 × 10-2 V
(9) This is an interesting problem – as well as being an illustration of Lenz’s Law, which
tells us that an induced current flows in such a direction that it opposes the change that
induced it. Here, a copper bar falls downward at a constant velocity, even though the
vertical conducting rails with which it is in contact are frictionless. This tells us that a
force equal and opposite to the weight force of the bar acts on it – therefore the bar is in
equilibrium, and the sum of the forces on it is zero.
The force is a magnetic force, exerted on it by a horizontal magnetic field. The bar
experiences this magnetic force because it is conducting a current – a current that is
induced because of the motion of the bar through the field. The bar moves downward,
and (in good Lenz’s Law style) the induced current flows through it in such a direction
that the resulting magnetic force is upward. Since the induced current, and therefore the
magnetic force on the bar, increases with increasing velocity, one can see that the bar will
at first accelerate with acceleration g, but as its velocity increases, the opposing magnetic
force also increases, so that the net downward force on the bar would decrease. When the
net force becomes zero, the bar no longer accelerates, falling instead at a constant
equilibrium velocity (which, in this problem, is 4.0 m/s).
Before we dive into the messy details of the solution to this particular problem, consider:
how would the world be different if Lenz’s Law worked in the opposite way? What
would be the consequences if induced currents flowed in such a direction as to reinforce
the change which caused them? Would this be a free-lunch, perpetual-motion, the-bigoil-companies-all-want-to-kill-us sort of universe?
Putting aside the perpetual motion, the motional EMF across the ends of the bar is:
E=vBL
(1)
Applying Ohm’s Law to the load resistance across the ends of the rails,
I=E/R=
LvB
.
R
(2)
The magnitude of the magnetic force on the bar is given by
Fm = ILB .
Substituting for I from equation (2),
vL2 B 2
⎛ LvB ⎞
.
Fm = ⎜
⎟ LB =
R
⎝ R ⎠
(3)
Since the bar descends at a constant velocity, it is in equilibrium, and the sum of the
forces on it must be zero. The only other force that acts on the bar is its weight, W = mg.
So, the equation of the bar’s equilibrium is
∑F
y
=0
Fm − W = 0
vL2 B 2
− mg = 0. Solving for m,
R
2
2
vL2 B 2 (4.0 m/s )(1.3 m ) (0.50 T )
m=
=
= 0.23 kg
Rg
(0.75 Ω ) 9.8 m/s 2
(
)
which is the part (a) answer. Part (b) asks us for the change in the bar’s gravitational
potential energy over an arbitrary short time interval of 0.20 s. (You can see right away
where they’re going with this.) The calculation is straightforward:
(
)
ΔGPE = mgΔh = mgvΔt = (0.23 kg ) 9.8 m/s 2 (− 4.0 m/s )(0.20 s ) = - 1.80 J
We called the 4.0 m/s negative, in this case, because the velocity is downward;
alternatively, we could simply have said that the change in gravitational potential energy
must be negative, since the descending bar is clearly losing gravitational potential energy.
In part (c), we’re asked for the energy dissipated in the load resistor during this same time
interval. In the absence of frictional losses, with energy being conserved, we would
expect that the “lost” gravitational potential energy of the bar should be dissipated as heat
in the resistor. But doing the calculation anyway, we call the energy dissipated by the
resistance E, and equate it to the product of power and time:
E = PΔt = I
2
2
2
(
[
(
LvB )
1.3 m )(4.0 m/s )(0.50 T )] ⋅ (0.20 s )
RΔt =
Δt =
= 1.80 J ,
just as we expected.
R
0.75 Ω
(13) This is one of those problems for which a picture is helpful.
θ
B
θ
B
The magnetic flux is given by
Φ = BA cosφ , where B is the field magnitude, A is the area, and φ is the angle between
the normal to the surface and the field vector. As the overhead-view picture above
shows, the initial position of the door (solid lines) places its surface perpendicular to the
field. There, its normal direction is parallel to the field, φ is zero, cos φ is one, and the
flux has its maximum value of BA. Note from the picture that as the door rotates
clockwise through an angle θ, the angle between its normal and the field direction is also
θ. Since B and A are both constant, cos φ ( = cos θ ) is the only thing that changes. So,
we want an angle θ such that cos θ = 1/3. That angle, arccos (1/3), is 70.5°.
(15) This time, I won’t draw a picture, since the authors of your textbook have provided
an exquisitely-shaded one on page 719. Theirs is vastly better than anything I would do.
Let us first reduce our workload by noting that three of the surfaces – the “bottom” and
the two triangular ends – are parallel to the field vector. Therefore their surface normals
are perpendicular to B, cos (90°) = 0, and the flux through each of these surfaces is zero.
Almost as easy is the “back” surface. It lies in the x-z plane, perpendicular to the field
direction. The flux through it is: Φ = BA = (0.25 T) (0.30 m) (1.2 m) = 0.09 Wb.
Finally, the slanting “roof.” The surface itself makes an angle with the field vector
whose tangent is 0.30 m / 0.40 m; this angle is arctan(3/4) = 36.87°. The normal to the
roof surface makes an angle (90° - 36.87°) = 53.13°. The flux is then
Φ = B A cos φ = (0.25 T) (1.20 m) (0.50 m) cos(53.13°) = 0.09 Wb.
This seems like a coincidence, until we consider that the “back” is simply the projection
of the “roof” on the X-Z plane anyway; the flux through both surfaces should be the
same.
Δ(BA cosφ )
ΔΦ
.
= −N
Δt
Δt
The magnetic flux through the coil changes because (in this case) the angle, φ, between
its normal and the magnetic field vector changes as the coil rotates in the field. We’re
given the induced EMF, the number of turns in the coil, the radius of the coil, and the
time interval during which its normal rotates from being perpendicular to the field, to
making an angle of 45° with the field. We can rewrite Faraday’s Law in a form that is
useful to us:
(19) This problem is solved using Faraday’s Law: EMF = − N
cosφ f − cosφ0
Δ cosφ
= − NBA
. Dropping the Lenz’s Law minus sign (we
Δt
Δt
are interested in the magnitude of B) and solving for B:
EMF = − NBA
B=
EMF ⋅ Δt
. Express the coil area in terms of its radius and substitute for A:
NA(cos φ f − φ0 )
B=
(0.065 V )(0.010 s )
EMF ⋅ Δt
=
= 8.56 × 10 -5 T
2
N πr (cos φ f − φ0 ) (950 ) π [0.060 m]2 (cos 45° − cos 90°)
( )
(
)
(25) As a coil is moved from a region of zero magnetic field to a region of an unknown
magnetic field, a current is induced in it, and the total amount of charge that moves
through the coil is measured. Our task is to calculate the magnitude of that field.
We note that the charge that flows is given by q = I Δt, where I is the average induced
current during the coil’s transition into the field. The induced current I, in turn, is given
by Ohm’s Law:
E
, where E is the induced EMF or voltage, and R is the coil’s resistance. This
R
leaves us to find the average induced EMF or voltage, which we get from Faraday’s Law:
I=
ΔΦ
. ΔΦ is the flux that we wish to measure (or, strictly speaking, it is the
Δt
difference between the flux we want to measure and zero). And, finally, B is related to Φ
E = −N
Φ
Φ
, which reduces to B =
since we are told that the coil remains
A cosφ
A
perpendicular to the magnetic field vector. Solving Faraday’s Law for Φ, which is the
same thing as ΔΦ since the initial field is zero, we have
by B =
Φ=
E ⋅ Δt
E ⋅ Δt
. Substituting into the equation above, we have B =
.
N
NA
Ohm’s Law allows us to substitute for E:
q
RΔt
IRΔt
qR
t
Δ
B=
, and the definition of current lets us substitute for I : B =
=
NA
NA
NA
Substituting numerical values from the problem (q = 8.87 x 10-3 C, R = 45.0 W,
N = 1850 turns, and A = 4.70 x 10-4 m2), we obtain: B = 0.459 T.
(31) Rather than copy the book’s artwork or draw my own, I’ll simply refer you to the
figure on p. 721 for this conceptual exercise. We’re looking down on the tabletop as the
loop is moved from above the wire to below, with the current going left-to-right in the
wire. Right-hand rule #2 tells us that the magnetic field vector points upward in the
region above the wire (position 1 in the figure), and downward in the region below the
wire (position 2). As the loop passes position 1, moving downward – and closer to the
wire – the upward flux through the loop is increasing (equation 21.5, p. 636, tells us that
B is inversely proportional to r). Lenz’s Law tells us that the induced current will flow in
a direction that opposes this change in flux – meaning that the induced current will
produce a downward flux. Right-hand rule #2 calls for a clockwise current to produce
that downward flux. So, as we pass position 1, we expect a clockwise induced current.
Passing position 2, still moving downward and away from the wire, we find that the
magnetic flux due to the current in the straight wire points downward through the loop …
and is also getting smaller (again, from equation 21.5, with r now increasing). Lenz’s
Law says that this change in flux – a decrease in downward flux – will induce a current
that tends to produce more downward flux, thus opposing the diminution in that flux. As
was the case when we passed position 1, a clockwise current will produce downward
flux, and so we again expect a clockwise induced current here.
(33) The current passing through the straight wire produces a magnetic field which
“wraps around” the wire, according to right-hand rule #2. The field vectors above the
wire point out of the page, and the vectors below the wire point down into the page.
Summed up over the entire loop, the net magnetic field is zero. As the current in the
straight wire increases, the field magnitudes above and below the straight wire also
increase, but the net increase, like the original net flux, is zero. Since ΔΦ is zero, no
current is induced in the circular loop.
(37) The peak EMF produced by a generator is given by EMFpeak = NABω . We’re
considering two generators, both using the same values of N and ω, but having different
values of A and B. We will call these values A1, A2, B1, and B2. All but B2 are given. B2
is a mystery. (Not much of a mystery, though, as we will see here.)
Since the peak EMFs are to be equal, we start out by writing
NA2 B2ω = NA1 B1ω . Dividing out N and ω, and solving for B2, we obtain
A1
0.045 m 2
B2 =
B1 =
⋅ 0.10 T = 0.30 T
A2
0.015 m 2
(45) From section 22.8 of your textbook, equation 22.11:
B2
. ρE (my notation -- your text does not provide one) is an energy density,
ρE =
2μ 0
meaning that it is the amount of energy stored in a unit volume of space that is occupied
by a uniform magnetic field of magnitude B. You can think of this as analogous to a
mass density, which is what we usually mean by ρ -- the amount of mass found in a unit
volume of space that is uniformly occupied by some material substance: air, water, rock,
or balsa wood. In any case, the energy found in a volume V, having an energy density
ρE, is simply
E = ρ EV =
B 2 ⋅ area ⋅ height (7.0 ×10 -5 T )2(5.0 ×108 m 2 )(1500 m )
=
= 1.5 ×10 9 J
-7
⋅
×
⋅
2μ 0
2 4π 10 T m/A
(48) Starting with equation 22.8 from your textbook: L =
NΦ
.
I
We know that the magnetic field inside a solenoid is parallel to its length, and
perpendicular to its cross-section. We can then substitute for Φ in the equation above:
L=
NA(μ 0 nI )
NBA
. In a solenoid, B = μ0 n I . Substituting for B, L =
= μ 0 nNA .
I
I
But the total number of turns in the coil is equal to the turns per unit length times the
length. Substituting for N, we have
L = μ 0 n 2 Al . Solving for n, we have n =
L
L
.
, and N = nl = l
μ 0 Al
μ 0 Al
Substituting numerical values from the problem: l = 0.052 m, L = 1.4 x 10-3 H, A = 1.2 x
10-3 m2 : N = 220 turns.
(57) The terms “step-up” and “step-down,” applied to transformers, refer to the
relationship between their input and output voltages. If we want to adapt our household
120 V service to operating a doorbell at 10 V, we clearly need a step-down transformer.
The turns ratio will be in proportion to the voltages:
N S VS 10.0 V
=
=
= 0.083 (that is, 1:12, secondary-to-primary).
N P VP 120 V
(59) We’re comparing the power lost to heating long-distance power transmission lines
for power transmitted at a high voltage (1200 V) to the loss if we transmit at very high
voltage (1.2 x 105 V, after a 100:1 step-up transformer). We first note that the total
resistance of the wires is the sum of the resistances of the two wires involved:
R = 2 (5.0 x 10-2 Ω/km) (7.0 km) = 0.70 Ω. This is true no matter what voltage we
choose for the transmission system.
Since we are told the amount of power that we must transmit in any case
(P = 1.2 x 106 W), we can write down the current that will flow, from the definition of
electrical power:
P = IV, or I =
P
.
V
P2R
The power dissipated as heat in the wires is given by PH = I R = 2 . Notice carefully
V
that P and PH, are very different things; PH is the power lost as heat in the wires, and P is
the power available for use at the “destination” end of our network. One might think of
the ratio P/PH as a measure of our transmission efficiency.
2
We’ve already established that R is 0.70 Ω, and P is 1.2 x 106 W. We now calculate the
power losses with and without the 100:1 step-up transformer.
V = 1200 V: PH = 7.0 x 105 W
V = 100 (1200 V) = 1.2 x 105 V: PH = 70W
We see, then, that by increasing the transmission voltage by a factor of 100, we reduce
the resistive transmission loss by a factor of 1002 or 10,000 -- in this example, from
enough power to operate about 5 typical houses to enough power to operate one bright
light bulb. Which is, of course, why utilities run high-voltage lines across country. The
actual power savings would be less, of course, due to losses in the step-up and step-down
transformers; but they will still be quite substantial.
(63) First, we are asked for the frequency of the generator. The frequency of any
periodic phenomenon of the reciprocal of its period, and we can obtain the period by
inspection of the plot of voltage vs. time given with the problem. There we see voltage
passing through zero in the positive direction at time t = 0, and we see this again at t =
0.42 s. The period T is then given by T = (0.42 s – 0 s) = 0.42 s. The frequency is then
f =
1
= 2.4 Hz
0.42 s.
Part (b) asks us for the generator’s angular speed:
ω = 2πf = 2π (2.4 Hz ) = 15 rad/s
Finally, part (c) requires the magnitude of the magnetic field in which the generator coil
rotates. Here, we note from the plot that Emax is 28 V; from equation 22.4,
E = NABω sin (ωt ) . Solving for B,
E
E
= max , since Emax occurs when ωt is 90° and its sine is one.
NAω sin (ωt ) NAω
Substituting the numerical values Emax = 28 V, N = 150 turns, A = 0.020 m2, and
ω = 15 rad/s, we have: B = 0.62 T.
B=
(65) In Figure 22.1 b, a south pole is approaching the end of the coil. To oppose this
change (as Lenz’s Law requires), the coil will have to behave as a magnet with its south
pole at that end. This means that the induced current must produce a field vector directed
into the coil from this end. Applying the right-hand rule (#2), the induced current in the
coil is clockwise as viewed from the end, so the current passes through the galvanometer
from right to left. In part c of the figure, a south pole is receding from the same end of
the coil. To oppose this motion, that end of the coil must act as a north pole, with B
protruding outward from the coil’s end. Again, right-hand rule #2 says that to do this, the
induced current flows counterclockwise as viewed from the end, and passes through the
galvanometer from left to right.