Chapter 33 Solutions
Exercise 1.
Model: Assume the magnetic field is uniform.
Visualize: The magnetic field is perpendicular to the axis of the wire so it's either into/out of the
page or up/down the page. Since a motional emf was developed the field must also be
r
perpendicular to v . The positive charges experienced ar magnetic force to the left. By the rightr
hand rule the field must be out of the page so that v × B is to the left.
Solve: This is a straightforward use of Eq. 33-3,
0.050 V
B= =
= 0.10 T , out of the page
vl (5.0 m/s )( 0.10 m )
Assess: This is reasonable, laboratory fields are typically up to a few tesla.
ε
Exercise 2.
Visualize: Take the antenna to be vertical and the car moving to the right (see figure). To
develop a motional emf the magnetic field needs to be perpendicular to both, so let's say into the
page.
Solve: This is a straightforward use of Eq. 33-3,
1.0 V
4
v=
=
= 2.0 × 10 m/s
-5
lB (1.0 m )(5 × 10 T )
ε
Assess: This is an unreasonable speed for a car. It's unlikely you'll ever develop a volt.
Exercise 3.
Visualize: The wire is pulled with a constant force in a magnetic field that results in a motional
emf and produces a current in the circuit. From energy conservation the mechanical power
provided by the puller must appear as electrical power in the circuit.
1
Solve: a) So we can use Eq. 33-6, P = Fpullv
b) We can again use Eq. 33-6, P =
⇒
v 2l 2 B 2
⇒B=
R
P
4.0 W
=
= 4.0 m/s
Fpull 1.0 N
v=
RFpull
vl
2
=
(0.20 Ω ) (1.0 N)
= 2.24 T
(4.0 m/s)(0.10 m )2
Assess: This is reasonable field.
Exercise 4.
Model: Assume the field changes abruptly at the boundary between the two sections.
Visualize: The flux is determined by how much field is passing through the loop. The
directions are opposite so some positive and some negative. The total flux is the sum of the
fluxes in the two regions.
r
Solve: The field is constant in each region so we use Eq. 33-10. Take A into the page, then it is
parallel to the field in the left region (#1) and the flux is positive, and opposite the field in the
right region
r r (#2)
r rso the
r flux
r will be negative.
Φ = A ⋅ B = A1 ⋅ B1 + A2 ⋅ B2 = A1 B1 cosθ + A2 B2 cosθ
Φ = (0.20 m)2 (2.0 T) - (0.20 m)2 (1.0 T) = 0.040 Tm 2 = 0.040 Wb
Assess: Overall positive since equal areas and the stronger field is parallel to the surface normal.
Exercise 5.
Model: Consider the solenoid to be long so the field is constant inside and zero outside.
Visualize: The field of a solenoid is along the axis. The flux through the loop is only non-zero
inside the solenoid. Since the loop completely surrounds the solenoid the total flux through the
loop will the same in both the perpendicular and tilted cases.
r
Solve: The field is constant inside ther solenoid so we use Eq. 33-10. Take A to be in the same
directionras ther field and
that B through the loop is zero outside the solenoid.
r recall
r
2
2
−5
a) Φ = Aloop ⋅ Bloop = Asol ⋅ Bsol = π rsol
Bsolcos
{ θ = π (0.010 m ) ( 0.20 T ) = 6.28 × 10 Wb
=1
2
r
Assess: When the loop is tilted the component of B in the direction of the loop normal is less,
but the effective area of the loop surface is increased by the same factor.
Exercise 6.
Model: Assume the field strength is uniform over the loop.
Visualize: According to Lenz' law the induced current creates an induced field that opposes the
change in flux.
Solve: The original field is into the page within the loop and is changing strength. The induced,
counterclockwise current produces a field out of the page within the loop that is opposing the
change. This implies that the original field must be increasing in strength so the flux into the
loop is increasing.
Exercise 7.
Model: Assume that the magnet is a bar magnet with field lines pointing away from the N end.
Visualize: As the magnets move, if they create a change in the flux through the solenoid there
will be an induced current and corresponding field. According to Lenz' law the induced current
creates an induced field that opposes the change in flux.
Solve: a) When magnet 1 is close to the solenoid there is some flux to the left through the
solenoid. As magnet 1 moves away there is less flux to the left. The induced current will oppose
this change and produce an induced current and a corresponding flux to the left. By the right
hand rule this corresponds to a current in the wire into the page at the top of the solenoid and out
of the page on the bottom. So, the current will be right to left across the resistor.
b) When magnet 2 is close to the solenoid the diverging field lines of the bar magnet produce a
flux to the left in the left half of the solenoid and a flux to the right in the right half. Since the
flux depends on the orientation of the loop, the flux on the two halves have opposite sign and the
net flux is zero. Moving the magnet away changes the strength of the field and flux, but the total
flux is still zero. Thus there is NO induced current.
Exercise 8.
Model: Assume plane of loop is perpendicular to the field direction.
Visualize: The flux is due to the field through the area of the triangle. Only the left half gives a
contribution as the field strength is zero on the right half.
r r
r
Solve: a) The
r flux is Φ = A ⋅ B . Take A into the page, perpendicular to the triangle, and thus
parallel to B , so in this case Φ = AB where A is the area of the half of the triangle where there is
a magnetic field. This smaller triangle has a base of 10 cm and height 20sin60° = 17.32 cm. Thus
Φ = AB = 12 (0.10 m)(0.1732 m)×0.1 T = 8.66×10–4 Wb.
3
b) The flux is directed into the loop. According to Lenz’s law, the induced current will try to
prevent the decrease of flux. To do this, the field of the induced current will have to point into
this loop. This requires the induced current to flow clockwise.
Exercise 9.
Visualize: The changing current in the solenoid produces a changing flux in the loop and by
Lenz's law there will be an induced current and field to oppose the change in flux.
Solve: The current shown produces a field to the right inside the solenoid. So there is flux to the
right through the surrounding loop. The current in the solenoid decreases there is less field and
less flux to the right through the loop. There is an induced current in the loop that will oppose
the change by creating an induced field and flux to the right. This requires a clockwise current.
Exercise 10.
Model: Assume the field is uniform.
Visualize: If the changing field produces a changing flux in the loop there will be a
corresponding induced emf and current.
Solve: The induced emf is ε = |dΦ/dt| and the induced current is I = ε/R. The field B is
changing, but the area A is not.
r
r
a) Take A out of the page and parallel to B , so Φ = AB and dΦ/dt = A(dB/dt). The field is
changing at the rate dB/dt = 0.5 T/s. Thus
ε = A dB = π r2 dB = π(0.050 m )2 (0.50 T/s) = 3.93 × 10 −3 V
dt
dt
3.93 × 10 −3 V
ε
I= =
= 3.93 × 10 −2 A = 39.3 mA
0.1Ω
R
The field is increasing out of the page. To prevent the increase, the induced field needs to point
into the page. Thus the induced current must flow clockwise.
b) As in part a, ε = A(dB/dt) = 3.93×10–3 V and I = 39.3 mA. Here the field is into the page and
decreasing. To prevent the decrease, the induced field needs to point into the page. Thus the
induced current must flow clockwise.
r r
r
r
c) Now A (left or right) is perpendicular to B and so A ⋅ B = 0 . That is, the field does not
penetrate the plane of the loop. If Φ = 0, then ε = |dΦ/dt| = 0 and I = 0. There is no induced
current.
Assess: Note that the induced field opposes the change.
Exercise 11.
4
Model: Assume the field strength is changing at a constant rate.
Visualize: The changing field produces a changing flux in the coil and there will be a
corresponding induced emf and current.
Solve: The induced emf of a coil is
ε
r r
d A⋅B
dΦ
dB
dB
=N
=N
= NA
= Nπ r 2
dt
dt
dt
dt
( )
r
r
where we’ve used the fact that B is parallel to A (see figure). With the parameters given,
ε = (10 3 turns)π (0.010 m )2 (0 − 0.10-3) T
= 3.14 V
10 × 10 s
The field is increasing out of the page. To prevent the increase, the induced field needs to point
into the page. Thus the induced current must flow clockwise.
Assess: This seems a reasonable emf as there are lots of turns.
Exercise 12.
Model: Assume the field is uniform.
Visualize: The motion of the loop changes the flux through it and this results in a corresponding
induced emf and current.
Solve: The induced emf is εr= |dΦ/dt| and the induced currentr is I = ε/R. The area A is changing,
but the field B is not. Take A out of the page and parallel to B , so Φ = AB and dΦ/dt = B(dA/dt).
Thus the flux is through that portion of the loop where there is field, A = lx. (See figure.)
5
ε = B dA
dt
=B
ε
d(lx )
= Blv = (0.20 T )(0.050 m )(50 m/s) = 0.50 V
dt
0.50 V
= 5.0 A
R 0.10Ω
The field is out of the page. As the loop moves the flux increases because more of the loop area
has field through it. To prevent the increase, the induced field needs to point into the page. Thus
the induced current must flow clockwise.
Assess: This seems reasonable since there is rapid motion of the loop.
I=
=
Exercise 13.
Model: Assume the field is uniform across the loop.
Visualize: There is a current in the loop so there must be an emf. That is due to a changing flux.
With the loop fixed the area is constant so the change in flux must be due to a changing field
strength.
Solve: The induced emf is ε = |dΦ/dt| rand the induced current is I = εr/R. The B field is
changing, but the area A is not. Take A into the page and parallel to B , so Φ = AB and dΦ/dt =
A(dB/dt).
ε = dΦ = d (AB) = A dB
dt
dt
dt
−3
dB IR (150 × 10 A )(0.10Ω )
=
=
= 2.34 T/s
R
R
(0.080 m )2
dt
A
The original field and flux is into the page. The induced counterclockwise current produces an
induced field and flux that is out of the page. Since the induced field opposes the change the
field must be increasing.
I=
ε = A dBdt
⇒
Exercise 14.
Model: The loops at different radii will be ignored and we assume the inductance is that of a
solenoid.
Visualize: The inductance of a solenoid depends on the number of turns, the length, and the
cross-sectional area. The length and area are determined from the given information. Let's find
the total number of turns and then the number of turns per layer. That along with the length will
give us the diameter of our wire.
Solve: The length in terms of the number of turns in a layer and the wire diameter is l = Nl dwire .
The total number of turns is N = 4N l . For a solenoid we have
6
Lsol =
µ 0N 2 A
l
⇒ N=
lLsol
=
µ0 A
(0.050 m )(100 × 10 −6 H)
= 225 turns
(4π × 10 −7 H/m )π (0.0050 m )2
Nl = N / 4 = 56, dwire = l / Nl = 0.0050 m / 56 = 8.9 × 10
−4
m = 0.89 mm
Exercise 15.
Model: Assume that the current drops uniformly.
Visualize: The changing current produces a changing flux in the inductor and a corresponding
emf is developed across the inductor.
Solve: For an inductor the potential difference is
dI
∆I
(50 − 150) × 10-3 A
−3
∆V = − L = − L
= −(10 × 10 H)
= 100 V
dt
∆t
(10.0 × 10 -6 s)
Assume that the current is flowing left to right as in the figure. As the current decreases so does
the flux and the induced current will oppose the change and try to maintain the original flux. This
means that the induced current will be in the same direction as the original current. If we
imagine the induced emf as a battery (inside dashed line) that has current flowing left to right
then the potential will increase in the direction of the current.
Assess: The inductor acts like a source of emf during the change.
Exercise 16.
Model: Assume that the current changes uniformly.
Visualize: We want to increase the current without exceeding a maximum potential difference.
Solve: Since we want the minimum time we will use the maximum potential difference
dI
∆I
∆I
(3.0 − 1.0 ) A
−3
−3
∆V = − L = − L
⇒ ∆t = L
= 200 × 10 H
= 1.00 × 10 s = 1.00 ms
dt
∆t
∆Vmax
(400 V )
Assess: If we change the current in any shorter time the potential difference will exceed the limit.
Exercise 17.
Visualize: The solenoid has inductance and when a current flows there is energy stored in the
magnetic field.
Solve: We can just use the energy for an inductor result once we know the inductance of the
solenoid.
7
2
2
−7
µ 0N 2 A (4π × 10 H/m )(200 ) π (0.015 m )
−4
Lsol =
=
= 2.96 × 10 H
(0.12 m)
l
U = 12 LI = 12 (2.96 × 10
2
−4
−5
H)(0.80 A) = 9.47 × 10 J
2
Exercise 18.
Visualize: Changing the variable capacitor in combination with a fixed inductor will change the
resonant frequency of the LC circuit.
Solve: Since the resonant frequency depends on the inverse square root of the capacitance the
lower capacitance will produce the higher frequency and vice versa.
1
1
1
6
ω=
⇒ ω max =
=
= 2.24 × 10 rad/s
−3
−1 2
LC
LCmin
2.0
×
10
H
100
×
10
F
(
)(
)
The corresponding minimum is ω min = 1.58 × 10 rad/s . These are angular frequencies so we can
ω
5
5
use f =
to find f min = 2.52 × 10 Hz and f max = 3.57 × 10 Hz .
2π
6
Exercise 19.
Visualize: A capacitor in combination with an inductor will have a resonant frequency
determined by the inductance and capacitance.
Solve: We know the frequency and the capacitance so we can find the inductance.
1
1
1
ω=
⇒ L= 2 =
2
LC
ω C (2πf ) C
=
1
(2π 100 × 10 Hz) (10 × 10
2
6
−1 2
F)
−7
= 2.53 × 10 H = 0.253 µH
Exercise 20.
Visualize: The rate at which the charges leave the capacitor will determine the current through
the inductor.
Solve: The charge on the capacitor is Q = Q0 cos ωt where Q0 = C∆V is the maximum charge at
dQ
= ωQ0 sin ωt . The sine function oscillates between +1 and
t=0. The current is given by I = −
dt
-1 so the maximum current is when the sine is unity.
 1 
C
0.10× 10 −6 F
−2
 C∆V = ∆V
Imax = ωQ0 = 
= (5.0 V )
= 5.00 × 10 A = 50 mA
 LC 
L
1.0 × 10 −3 H
Exercise 21.
8
Visualize: This is a simple LR circuit if the resistors in parallel are treated as an equivalent
resistor in series with the inductor.
Solve: We can find the equivalent resistance from the time constant as we know the inductance.
L
L
L 3.6 × 10−3 H
τ= =
⇒ Req = =
= 360Ω .
τ
10 × 10 −6 s
R Req
The equivalent resistance is the parallel addition with the unknown as R2 and R1=600 Ω.
R1Req
1
(600Ω )(360 Ω)
1
1
=
+
⇒ R2 =
=
= 900Ω
R1 − Req (600 Ω) − (360Ω )
Req R1 R2
Exercise 22.
Visualize: This is a simple LR circuit if the two resistors in series are treated as an equivalent
resistor in series with the inductor.
Solve: First we find the time constant of the equivalent circuit knowing the inductance and
capacitance.
L
L
50 × 10 −3 H
−4
τ=
=
=
= 1.00 × 10 s
Req R1 + R2
500Ω
If the current is initially at the maximum then it will decay exponentially with time,
−t/ τ
I(t) = I0 e
. We can find the time, t1/ 2 , when it drops to half the initial value.
−t
/τ
1
⇒ ln (12 ) = −t1 / 2 / τ ⇒ t1/ 2 = τ ln 2 .
2 I0 = I0 e
This result is quite general and is characteristic of all exponential decay phenomena. So we find
−4
−5
t1/ 2 = τ ln 2 = (1.00 × 10 s)0.693 = 6.93 × 10 s
1/ 2
Assess: This is less than τ , as expected, since the time constant is the amount of time it takes to
decay to 0.37 of the initial current and we are only going to 1/2.
Problem 23.
Visualize: To calculate the flux we need to consider the orientation of the surface normal
relative to the magnetic field direction. We will consider the flux through the surface in two
parts corresponding to the different surface normal directions. The flux through the parts will be
added to get the total flux.
9
Solve: The figure shows an edge view of the loop. The flux when the field is uniform is
r
r r
r
Φ = Φ top + Φleft = Atop ⋅ B + Aleft ⋅ B
= AtopBcos 45 + Aleft B cos45
o
o
= (0.050 m × 0.10 m)(0.050 T)cos 45 +
o
(0.050 m × 0.10 m)(0.050 T) cos45o
= 3.54 × 10−4 Wb
Problem 24.
Model: Assume the field is uniform inside the rectangular region and zero outside.
Visualize: The flux measures how much of the field penetrates the chosen surface. We will
break the surface up as the magnetic field has different strengths over different parts of the
surface of the loop.
Solve: For convenience we choose the normal to the loop into the page so it's in the same
direction as the magnetic field.
r r
r r
r r
r r
Φ = ∫ B ⋅ dA = ∫ B ⋅dA + ∫ B ⋅ dA = ∫ B ⋅ dA + 0
loop
=
inside
rectangle
outside
rectangle
inside
rectangle
∫ BdA cosθ = ∫ BdA = B ∫ dA = BA
inside
rectangle
inside
rectangle
inside
rectangle
inside
rectangle
= Bab
Assess: The flux depends directly on the strength of the field and the size of the loop.
Problem 25.
Model: Assume the field is uniform in space though it is changing in time.
10
Visualize: The changing magnetic field strength produces a changing flux through the loop, and
a corresponding induced emf and current.
Solve:
r
r
a) Since the field is perpendicular to the plane of the loop A is parallel to B , so Φ = AB
2
and ε = dΦ / dt = A dB/ dt . Since B = 4t − 2t , we find dB / dt = 4 − 4t T/s. If the square has
sides of length b, then
ε = A dB
dt
⇒
I=
= b2 4 − 4t = (0.20 m)2 (4 − 4t )T/s = 0.16(1 − t ) V
ε
R
=
0.16(1 − t ) V
= 1.6(1 − t ) A
0.10Ω
We can now tabulate B, ε, and I at the different times.
time (s)
B (T)
ε (volts)
I (A)
0.0
0.00
0.16
1.6
0.5
1.50
0.08
0.8
1.0
2.00
0.00
0.0
1.5
1.50
-0.08
-0.8
2.0
0.00
-0.16
-1.6
Using the formulas obtained above we determine the magnitude of the emf and the current.
However, the magnetic field is increasing over the interval 0 < t < 1 s and decreasing over the
interval 1 s < t < 2 s so the induced emf and current must have the opposite sign in the second
half of the time interval. We arbitrarily choose it positive during the first half.
b)
To plot the field and current we look at the form of the equations as a function of time.
The magnetic field strength is quadratic with a maximum at t =1 s and vanishing at t = 0, 2 s.
The current equation is linear and decreasing starting at 1.6 A at t =0 s and going through zero at
t = 1 s.
11
Assess: Notice in the graph how I = 0 at t = 1 s, the instant in time when B is a maximum, dB/dt
= 0, and thus the flux is (instantaneously) not changing so the corresponding induced emf and
current are zero.
Problem 26.
Model: Assume the field is uniform in space though it is changing in time.
Visualize: The changing magnetic field strength produces a changing flux through the coil, and
a corresponding induced emf and current.
Solve:
2
a) B = 0.020t + 0.010t
r
r
b)
Since the field is perpendicular to the plane of the coil A is parallel to B , so Φ = AB
2
and ε = N dΦ / dt = NA dB / dt . Since B = 0.020t + 0.010t , we find
dB / dt = (0.020 + 0.020t ) T/s . The magnetic field is always positive and increasing so we will
drop the absolute value in what follows.
ε (t) = NA dB
dt
= Nπ r ( 0.020 + 0.020t )
2
= 20π (0.025 m)2 (0.020 + 0.020t )T/s = 7.85 × 10 −4 (1 + t) V
⇒
−4
7.85 × 10 (1+ t) V
ε
(t)
I(t) =
=
= 1.57 × 10 −3 (1 + t) A
R
0.50Ω
c)
Using our expression for I(t) we can just plug in and get
−3
−3
−2
I(5 s) = 1.57 × 10 (1 + 5) A = 9.42 × 10 A and I(10 s) = 1.73 × 10 A
Assess: The field is always increasing and increasing more rapidly with time so the induced
current is greater at 10 s than at 5 s.
Problem 27.
Model: Assume the field is uniform in space though it is changing in time.
12
Visualize: The changing magnetic field strength produces a changing flux through the coil, and
a corresponding induced emf and current.
Solve:
2
a) B = t − 14 t
r
r
b)
Since the field is perpendicular to the plane of the coil A is parallel to B , so Φ = AB
2
and ε = N dΦ / dt = NA dB / dt . Since B = t − 14 t , we find dB / dt = (1 − 12 t ) T/s .
ε (t) = NA dB
dt
= Nπ r (1− 12 t ) T/s
2
= 50π (0.020 m) (1 − 12 t )T/s = 6.28 × 10 (1 − 12 t) V
−2
2
⇒
I(t) =
ε (t) = 6.28 × 10 −2 (1− 12 t) V = 6.28 × 10 −2 (1 − 1 t) A
R
1.0Ω
2
c)
Using our expression for I(t) we can just plug in and get
−2
−2
I(1 s) = 6.28 × 10 (1 − 12 1) A = 3.14 × 10 A = 31.4 mA , we also get I(2 s) = 0.0 A
I(3 s) = −31.4 mA
Problem 28.
Model: Assume that B changes uniformly with time.
Visualize: The magnetic strength is changing so the flux is changing and this will create an
induced emf.
13
Solve: The magnetic field is at an angle θ
= 60° to the normal of the plane of the coils, so the
r r
flux for a single loop of the coil is Φ = A ⋅ B = BA cos θ . The radius and angle don’t change with
time, but B does. According to Faraday’s law, the induced emf is
ε (t) = N dΦ = NAcos θ dB = Nπ r2 cos θ ∆B
dt
∆t
dt
= 100π (0.010 m) cos60
2
(1.50 − 0.50) T
0.60 s
−2
= 2.62 × 10 V
where we have included the number of turns in the coil and used
dB ∆B
=
.
dt
∆t
Problem 29.
Model: Assume that B changes uniformly with time.
Visualize: The magnetic strength is changing so the flux is changing and this will create an
induced emf.
Solve: The magnetic field is at an angle θ
= 45° to the normal to the plane of the coils, so the
r r
flux for a single loop of the coil is Φ = A ⋅ B = BA cos θ . The radius and angle don’t change with
time, but B does. According to Faraday’s law, the induced emf is
ε (t) = N dΦ = NAcos θ dB = Nπ r2 cos θ ∆B
dt
∆t
dt
= 25π (0.050 m) cos45
2
(0.20 − 0.80 ) T
2.0 s
= 4.17 × 10
where we have included the number of turns in the coil and used
14
−2
V = 41.7 mV
dB ∆B
=
.
dt
∆t
Problem 30.
Model: Assume that the field is uniform in space over the coil.
Visualize: We want an induced current so there must be an induced emf created by a changing
flux. If we assume that the coil doesn't change size or orientation then the magnetic field
strength must be changing.
Solve: To relate the emf and the current we need to know the resistance. We have that
ρ l
2
R = C u wire . We use the given size of copper wire to get the area, Awire = π rwire
. From the size
Awire
of the coil and the number of turns we can get the length of the wire, lwire = N2πr so
−8
ρ C ulwire ρ C uN2π r 2Nρ C ur 2(100)(1.7 × 10 Ω m )(0.040 m )
R=
=
=
=
= 2.18Ω
2
2
2
rwire
Awire
πrwire
(2.5 × 10 −4 m)
The magnetic
r r field is perpendicular to the plane of the coil so the flux for a single loop of the coil
is Φ = A ⋅ B = BA , if we take the normal to the coil to be in the same direction as the field.
According to Faraday’s law, the induced emf and current are
dB
ε
dB
= Nπ r 2
and I =
R
dt
dt
dB
(2.0 A )(2.18Ω)
IR
=
= 8.67 T/s
2 =
100π ( 0.040 m )2
dt
Nπ r
where we have included the number of turns in the coil.
ε
= NA
Problem 31.
Model: Assume the wire is long enough so we can use the magnetic field of an “infinite” wire.
15
Visualize: The magnetic field in the vicinity of the loop is due to the current in the wire and is
perpendicular to the loop. The current is changing so the field and the flux through the loop are
changing. This will create an induced emf and induced current in the loop.
Solve: The induced current depends on the induced emf
ε
1 dΦ
Iloop = loop =
R
R dt
We need to be careful when determining the flux since the field strength is not the same over all
parts of the loop. The flux through a rectangular loop due to a wire was found in Example 33-5.
The flux through a small strip of width dx is
r
r
µ Ibdx
dΦ = Bwire ⋅ dA = Bwire dA = Bwire (bdx) = 0
2πx
where we used Bwire = µ0I/2π x at a distance x from a wire. Integrating from the near edge, at x =
c, to the far edge, at x = c + a, gives the total flux:
µ Ib c +a dx µ 0 Ib  c + a 

Φ = 0 ∫c
=
ln
2π
x
2π  c 
For this problem a = b = 0.020 m so the induced current is
Iloop =
1 d  µ0Ib  c + b   1 µ0b  c + b  dI

ln
ln
 =

R dt  2π  c   R 2π  c  dt
(4π × 10
=
T m/A ) (0.020 m )  0.030 m 
−5
 100 A/s = 4.39 × 10 A = 43.9µA
ln
 0.010 m 
(0.010 Ω )2π
−7
Problem 32.
Model: Since the solenoid is fairly long compared to its diameter and the loop is located near
the center we will take the solenoid field to be uniform inside and zero outside.
Visualize: The solenoid’s magnetic field is perpendicular to the loop and creates a flux through
the loop. This flux changes as the solenoid's current changes, causing an induced emf and
corresponding induced current.
Solve: The induced current from the induced emf is given by Faraday's law
ε
1 dΦ
Iloop = loop =
R dt
R
2
The field is ≈0 outside the solenoid, so the flux is confined to the area Asol = π rsol of the solenoid,
not the larger area of the loop itself. Thus
16
dBsol
π r 2 µ N dI
2 d  µ0 NI 
= π rsol

 = sol 0
dt  l 
l
dt
dt
where Bsol = µ0NI/l. The current is constant from 0 s ≤ t ≤ 1 s and 2 s ≤ t ≤ 3 s, so dI/dt = 0 and
Iloop = 0 during these intervals. The current is changing at the rate |dI/dt| = 40 A/s during the
interval 1 s ≤ t ≤ 2 s, so the induced emf and current during this interval are
2
−7
2
π rsol
µ0 N dI π (0.010 m ) (4π × 10 T m/A )100
-5
ε loop =
=
40 A/s = 1.58 × 10 V
0.10 m
l
dt
εloop 1.58 × 10-5 V
Iloop =
=
= 1.58 × 10-4 A = 158µA
0.10Ω
R
During the first half of this interval, from 1.0 s to 1.5 s, the field is to the right and decreasing. To
try preventing the decrease, the induced field must point to the right. This requires an induced
current coming out of the page at the top of the loop – a positive current with the sign definition
given in the problem. During the second half of the interval, from 1.5 s to 2.0 s, the field is to the
left and increasing. To try preventing the increase, the induced field must point to the left. This
also requires a positive induced current coming out of the page at the top of the loop. Thus
ε
= Asol
 0
0 s ≤ t ≤ 1s

Iloop = 158 µA 1 s ≤ t ≤ 2s

 0
2 s ≤ t ≤ 3s
This information is shown in the graph.
Assess: The induced current is essentially the derivative of the solenoid current.
Problem 33.
Model: Assume the bottom of the loop is on the x axis.
Visualize: The field is changing with time so the changing flux will produce an induced emf and
corresponding induced current in the loop. The field strength varies in space as well as time so
we need to be careful when determining the flux.
17
Solve: Let's take the surface normal of the loop to be in the z direction so it is parallel to the
magnetic field. The flux throughrthershaded strip is given by
2
dΦ = B ⋅ dA = B dA = B(bdy) = (0.80y t )bdy .
To find the total flux we integrate over the area of the loop
Φ = ∫ (0.80y t )bdy = 0.80tb∫ y dy = 0.80tb / 3
b
b
2
0
2
4
0
To find the induced emf and current we take the time derivative of the flux
εloop 1 dΦ 1 d  0.80tb 4  0.80b 4 0.80(0.20 m )4
−4
Iloop =
=
=
=
= 8.53 × 10 A

=
R
3R
3(0.50Ω )
R dt
R dt  3 
Problem 34.
Model: Since the solenoid is fairly long compared to its diameter and the coil is located near the
center we will take the solenoid field to be uniform inside and zero outside.
Visualize: The solenoid’s magnetic field is parallel to the coil's axis and creates a flux through
the coil. This flux changes as the solenoid's current changes, causing an induced emf and
corresponding induced current in the coil.
Solve: The induced current from the induced emf is given by Faraday's law
1
ε
dΦcoil
Icoil = coil = Ncoil
R
R
dt
The field is constant inside the solenoid and through the loops of the coil, the coil flux is
2
confined to the area Acoil = π rcoil of the coil, not the larger area of the solenoid. Thus
ε
1
1
N π r 2 µ N dI
dB
d  µ0 NI 
2

 = coil coil 0
Icoil = coil = Ncoil Acoil sol = Ncoilπ rcoil
R
R
dt  l 
Rl
dt
R
dt
where Bsol = µ0NI/l. The current is changing uniformly over the interval 0 s ≤ t ≤ 0.02 s at the
rate |dI/dt| = 50 A/s so the induced current during this interval is
2
−7
2
µ0 N dI 5π (0.0050 m ) (4π × 10 T m/A )(120)
Ncoilπ rcoil
−4
Icoil =
=
50 A/s = 3.7 × 10 A
(0.1Ω)(0.080 m )
Rl
dt
The current in the solenoid is changing steadily so the induced current in the coil is constant. The
current is initially positive so the field is initially to the right and decreasing. The induced
18
current will try to oppose this change and will therefore produce an induced field to the right.
This requires an induced current in the coil coming out of the page at the top, so it is also
positive.
Problem 35.
Model: We assume that the magnetic field of coil 1 passes through coil 2 and that we can use
the magnetic field of a solenoid for coil 1.
Visualize: The field of coil 1 produces a flux in coil 2. The changing current in coil 1 gives a
changing flux in coil 2 and a corresponding induced emf and current in coil 2.
Solve: The induced current from the induced emf is given by Faraday's law
ε 1 dΦ 2
I2 = 2 = N2
R R
dt
The field of coil 1 is constant inside the loops of coil 2, the flux is confined to the area
2
A2 = πr2 of coil 2. Thus
2
ε 1
dB
d µ NI 
1
N π r µ N dI
I2 = 2 = N2 A2 1 = N2π r22  0 1 1  = 2 2 0 1 1
dt  l1 
R
R R
dt
Rl1
dt
20π ( 0.010 m ) (4π × 10 T m/A )( 20)
20 A/s = 7.95 × 10−5 A = 79µA
(2Ω)(0.020 m )
where B1 = µ0 N1 I1 / l1 is the field of coil 1, and l1 = N1d = 20(1.0 mm ) = 0.020 m . The current is
changing uniformly over the interval 0.10 s ≤ t ≤ 0.30 s at the rate|dI/dt| = 20 A/s.
2
−7
=
The current in coil 1 is changing steadily over the time interval so the induced current in the coil
2 is constant. From 0 s to 0.1 s and 0.3 s to 0.4 s current 1 is constant so current 2 vanishes.
From 0.1 s to 0.2 s the current in coil 1 is initially negative so the field is initially to the right and
the flux is decreasing. The induced current will try to oppose this change and will therefore
produce a field to the right. This requires an induced current in coil 2 that comes out of the page
at the top of the loops so it is negative. From 0.2 s to 0.3 s the current in coil 1is positive so the
field is to the left and the flux is increasing. The induced current will try to oppose this change
and will therefore produce a field to the right, again this is a negative current.
19
Assess: We see that the induced current in coil 2 is essentially the time derivative of the current
in coil 1.
Problem 36.
Model: Assume the magnetic field is uniform over the plane of the loop.
Visualize: The oscillating magnetic field strength produces a changing flux through the loop and
an induced emf in the loop.
Solve: a) The magnetic field is given by B = B0 sin ω t = (20 nT ) sin ωt . Let's choose the surface
r r
normal of the loop to be in the same direction as the magnetic field is initially Φ = A ⋅ B = BA .
The induced emf is
ε = dΦ = A dB = π r 2 dB = π r2ωB0 cosω t
dt
dt
dt
The cosine will oscillate between +1 and -1 so the maximum emf will be
ε max = π r 2ωB0 = π r2 (2πf )B0 = 2π 2r2 fB0 = 2π 2(0.125)2 (150 × 106 Hz)20 × 10 −9 T = 0.925 V
b) If the loop is rotated so that the plane is perpendicular to the electric field then the surface
normal is perpendicular to the magnetic field. There is no magnetic flux through the loop and no
induced emf. ε max = 0 V .
Problem 37.
Model: Assume the field due to the solenoid is uniform inside and vanishes outside.
Visualize: The changing current in the solenoid produces a changing field and flux through the
coil. This changing flux creates an induced emf in the coil.
Solve: The flux isr only
r non-zero within the area of the solenoid, not the entire area of the coil.
The flux is Φ = A ⋅ B = BA and the induced current in the coil is given by
20
Icoil =
=
ε coil
R
=
2


dΦ 1
dB
1
1
2 d
 µ0 Ns Is  = Ncoilπ rs µ0 Ns dIs
Ncoil
= Ncoil As sol = N coilπ rs
dt
dt  ls 
dt
Rls
dt
R
R
R
N coilπ rs2 µ0 Ns
2πf I0 cos(2π ft )
Rls
50π (0.010 m ) (4π × 10 −7 T m/A )200
2
=
(0.50Ω)(0.20 m )
= ( 7.44 mA )cos(2π ft )
2π (60 Hz )( 0.50 A ) cos(2π ft )
Assess: The induced current oscillates because the current and field of the solenoid oscillate.
Problem 38.
Model: Assume the field due to the solenoid is uniform inside and vanishes outside.
Visualize: The changing current in the solenoid produces a changing field and flux through the
coil. This changing flux creates an induced emf in the coil.
Solve: The flux is only non-zero within the area of the solenoid, not the entire area of the coil.
The induced current in the coil is given by
ε
1
1
N π r2 µ N dI
dΦ 1
dB
d µ N I 
Icoil = coil = Ncoil
= Ncoil As s = Ncoilπ rs2  0 s s  = coil s 0 s s
R
R
R
dt
dt  ls 
R
dt
Rls
dt
N coilπ rs2 µ0 Ns
2πf I0 cos(2π ft )
Rls
The cosine has extrema of +1 and -1 so the maximum coil current is
N π r 2µ N
Icoil,max = coil s 0 s 2π f I0
Rls
which gives
Icoil,max Rls
(0.20 A )(0.40Ω )(0.20 m )
I0 =
=
= 5.97 A
2
Ncoilπ rs µ0 Ns 2πf
40π (0.015 m )2 (4π × 10 −7 T m/A )(200)2π (60 Hz )
=
Assess: This is a large current, but the induced current is large as well.
Problem 39.
Model: Assume an ideal transformer.
Visualize: An ideal transformer changes the voltage, but not the power (energy conservation).
Solve: a) The primary and secondary voltages are related by Eqn. 33-23,
N
V
15,000
V2 = 2 V1 ⇒ N1 = 1 N2 =
100 = 12,500
N1
V2
120
21
b) The input power equals the output power and we recall that P = I∆V , so
I ∆V
(250 A )120 V
Pout = Pin ⇒ I1∆V1 = I2 ∆V2 ⇒ I1 = 2 2 =
= 2.0 A
15,000 V
∆V1
Assess: These seem reasonable, houses have low voltage and high current while the
transmission lines have high voltage and low current.
Problem 40.
Model: Assume the field changes abruptly at the boundary and is uniform.
Visualize: As the loop enters the field region the amount of flux will change as more area has
field penetrating it. This change in flux will create an induced emf and corresponding current.
While the loop is moving at constant speed the rate of change of the area is not constant because
of the orientation of the loop. Care must be taken in determining the change in flux.
Solve: a) First we determine how the position of the loop is related to the flux through it and
then find the induced current by the time derivative. If the edge of the loop enters the field
region at t = 0 s, then the leading corner has moved a distance x = v0 t . The area of the loop with
flux through it is A = 2(12 )yx = x 2 = (v 0t ) where we have used the fact that y = x since the sides
of the loop arer oriented
at 45o to the horizontal. Take the surface normal of the loop into the
r
page so Φ = A ⋅ B = BA .
 2(0.80 T )(10 m/s)2 
ε 1 dΦ 1 dA 1 d
2
1
2

 t = 1.6 × 10 3 t A .
I= =
= B
= B (v 0t ) = B2v0t = 
R R dt
0.10 Ω
R dt
R dt
R


The current is increasing at a contant rate. This expression is good until the loop is halfway in.
−3
The time for the loop to be halfway can be found. b / 2 = v0t = (10 m/s )t ⇒ t = 7.07 × 10 s
at which time the current is 11.3 A. While the second half of the loop is moving into the field
the flux continues to increase, but at a slower rate. Therefore, the current will decrease at the
same rate as it increased before, until the loop is completely in the field. After that the flux will
not change and the current will be zero.
2
22
b) The maximum current occurs when the flux is changing the fastest and this occurs when the
loop is halfway.
Problem 41.
Model: The inner loop is small enough that the field due to the outer loop is essentially uniform
over its area. Also, take the current in the outer loop to change uniformly.
Visualize: The current in the outer loop creates a field at the location of the inner loop.
Changing the current in the outer loop results in a change in flux and corresponding induced emf
and current in the inner loop.
Solve: The
field due to the outer loop is perpendicular to the plane of the little loop so
r magnetic
r
Φ inner = Ainner ⋅ Bouter = Ainner Bouter . The radius and angle don’t change with time, but Bouter does.
According to Faraday’s law, the induced emf is
Iinner =
ε inner(t)
Rinner
=
2
2
2
π rinner
µ0 dIouter
π rinner
µ 0 ∆Iouter
1 dΦ π rinner d  µ0 Iouter 

 =
=
=
Rinner dt  2router  2Rinnerrouter dt
2Rinnerrouter ∆t
Rinner dt
π (0.0010 m) 2(4π × 10 −7 T m/A ) (−1.0 − 1.0) T
=
= 3.95 × 10 −8 A = 39.5 nA
0.10 s
2(0.02Ω)(0.050 m )
dI ∆I
=
where we used
.
dt ∆t
Problem 42.
Visualize: The moving wire in a magnetic field results in a motional emf and a current in the
loop.
Solve: a) The induced emf is εloop = vlB = (100 m/s)(0.040 m)(1.0 T) = 4.0 V.
b) The induced current is I = εloop/R = (4.0 V)/(0.040 Ω) = 100 A.
c) The slide wire generates the emf , εloop. However, the slide wire also has an “internal
resistance” due to the resistance of the slide wire itself – namely r = 0.01 Ω for one-fourth of the
23
square. Thus the slide wire acts like a battery with an internal resistance. The potential difference
between the ends is
∆V = εloop – Ir = 4.0 V – (100 A)(0.01 Ω) = 3.0 V.
It is this 3.0 V that drives the 100 A current through the remaining 0.03 Ω resistance of the
“external circuit.”
Assess: The slide wire system acts as an emf just like a battery does.
Problem 43.
Model: Assume that the field is uniform in space over the loop.
Visualize: The moving slide wire in a magnetic field develops a motional emf and
corresponding current in the wire and the rails. The current through the resistor will cause
energy dissipation and subsequent warming.
Solve: a) The induced emf depends on the changing flux, not on where the resistance in the
loop is located. So we have
εloop = vlB = (10 m/s)(0.20 m)(0.10 T) = 0.20 V.
The total resistance around the loop is R = 1 Ω, due entirely to the carbon resistor, so the induced
current is
b) To move with constant velocity the acceleration must be zero, so the pulling force must
balance the retarding magnetic force on the induced current in the slide wire. Thus
Fpull = Fmag = IlB = (0.20 A)(0.20 m)(0.1 T) = 4.0×10–3 N.
c) Current in the resistor will result in power dissipation
P = I2R = (0.20 A)2(1 Ω) = 0.040 W = 0.040 J/s.
During a 10 second period Q = 0.40 J of energy are dissipated by the current, increasing the
internal energy and raising the temperature of the carbon. This is, effectively, a heat source. The
heat is related to the temperature rise by Q = mc∆T, where c is the specific heat of carbon. Thus
∆T =
Q
0.40 J
o
=
= 11 C
−5
o
mc (5.0 × 10 kg)(710 J/kg C)
Assess: This is a significant change in the temperature of the resistor.
24
Problem 44.
Model: Assume no resistance in the rails. If there is any, it is accounted for by the resistor.
Visualize: The moving wire will have a motional emf that produces a current in the loop.
Solve: a) At constant velocity the external pushing force is balanced by the magnetic force so
 Blv 
B 2l 2 v (0.50 T )2 (0.10 m )2 (0.50 m/s )
−4
Fpush = Fmag = IlB = 
=
= 6.25 × 10 N
 lB =
 R 
2.0 Ω
R
ε
1
where we used I = = Blv .
R R
b) The power is P = Fv =
B 2 l 2v 2 (0.50 T )2 (0.10 m )2( 0.50 m/s)2
−4
=
= 3.13 × 10 W
R
2.0Ω
c) Flux is out of the page and decreasing so the induced current/field will oppose the change.
The induced field must have flux that is out of the page so the current will be counterclockwise.
d) To find the power dissipated by the resistor we need the current, which is given by
 Blv  ( 0.50 T )(0.10 m)(0.50 m/s )
I=
= 1.25 × 10 −2 A
=
 R 
2.0Ω
Now we can determine the power
2
2
−2
−4
P = I R = (1.25 × 10 A ) ( 2.0 Ω) = 3.13 × 10 W
Assess: From energy conservation we see that the mechanical energy put in by the pushing force
shows up a electrical energy in the resistor.
Problem 45.
Model: Assume that the magnetic field is uniform in the region of the loop.
Visualize: The rotating semicircle will change the area of the loop and therefore the flux
through the loop. This changing flux will produce an induced emf and corresponding current in
the bulb.
Solve: a) The spinning semicircle has a surface normal that changes in time, so while the
magnetic field is constant, the area is changing. The flux through in the lower portion of the
circuit does not change and will not contribute to the emf. Only the flux in the part of the loop
containing
r r the rotating semicircle will change. The flux associated with the semicircle is
Φ = A ⋅ B = BA = BA cosθ = BA cos(2π ft ) , with θ the angle between the normal of the rotating
semicircle and the magnetic field, we have used θ = ω t , and A is the area of the semicircle. The
induced current from the induced emf is given by Faraday's law
25
I=
ε
R
=
1 dΦ 1 d
B πr2
=
BA cos(2πft ) =
2π f sin(2π ft )
R dt
R dt
R 2
2( 0.20 T )π ( 0.050 m )
f sin(2πft ) = 4.93 × 10 −3 f sin(2π ft ) A s
2(1.0Ω )
b) Now we can solve for the frequency necessary to achieve a certain current. From dc circuits
2
we know how power relates to resistance so P = I R ⇒ I = P / R = 4.0 W/1.0Ω = 2.0 A .
The sine function is max of 1 so we have
2.0 A
Imax = 4.93 × 10 −3 f A s = 2.0 A ⇒ f =
= 405 Hz
4.93 × 10 −3 A s
=
2
2
Assess: This is not reasonable by hand.
Problem 46.
Model: Assume the magnetic field is uniform over region where the bar is sliding and that
friction between the bar and the rails is zero.
Visualize: The battery will produce a current in the rails and bar. This current-carrying bar will
experience a force. With the battery as shown the current will be down in the bar and by the
right hand rule the force on the bar will be to the right. The motion of the bar will change the
flux through the loop and there will be an induced emf that opposes the change.
Solve: a) & b) As the bar speeds up the induced emf will get larger until finally it equals the
battery emf. The current will go to zero and the bar will continue at constant velocity.
1.0 V
ε = Blv term = ε bat ⇒ v term = ε bat =
= 33.3 m/s
Bl
(0.50 T )(0.060 m )
Assess: This is pretty fast.
Problem 47.
Model: Assume the magnetic field is uniform in the region of the loop.
Visualize: The moving wire creates a changing area and corresponding change in flux. This
produces an induced emf and induced current. The flux through the loop depends on the size and
orientation of the loop.
26
r r
Solve: a) The surface normal is perpendicular to the loop and the flux is Φ inner = A ⋅ B = AB cosθ ,
we can get the current from Faraday's law
ε 1 dΦ 1 d
B cosθ d
Blv cos θ
I= =
=
AB cosθ =
lx =
R dt
R R dt
R dt
R
b) From the free-body diagram shown we use Newton's second law. The magnetic force on a
straight, current-carrying wire, Fm = IlB is horizontal. Using the current I from above gives
B2 l 2 v cos2 θ
+ mgsin θ = max
∑ Fx = − Fm cosθ + mgsin θ = −
R
The acceleration will be zero when the two terms are equal, then v = v term
B 2l 2 vcos 2 θ
mgRtan θ
+ mgsinθ = 0 ⇒ v term = 2 2
R
l B cosθ
Assess: This seems reasonable.
−
Problem 48.
Model: Assume the magnetic field is uniform and that the change occurs abruptly.
Visualize: The figure shows the front of the loop, with the field coming out of the page. As the
loop falls, the flux will decrease. This will induce a current to oppose the decrease, so the
induced current will flow counterclockwise. As this current passes through the top of loop, it
experiences an upward magnetic force. There is no force on the bottom of the loop because B= 0
at the bottom. So there is an upward force – a retarding force – on the loop as it falls in the field.
As the loop speeds up the retarding force will become larger until it balances the weight.
27
Solve: a) The force on the current in the top arm of the loop is Fm = IlB. The forces on the side
arms of the loop cancel. The induced current is
ε 1 dΦ 1 d
B d
Blv
I= =
=
AB =
lx =
R R dt
R dt
R dt
R
Consequently, the retarding magnetic force is
l 2B 2 v  l 2B 2 
 v
Fm =
= 
 R 
R
The important point is that Fm is proportional to the speed v. As the loop begins to fall and its
speed increases, so does the retarding force. Within a very short time, Fm will increase in size to
where it matches the downward weight force w = mg. At that point, there is no net force on the
loop, so it will continue to fall at constant speed.
The condition that the upward magnetic force magnitude equal the weight magnitude is
2
 l2B2 
mgR (0.010 kg)(9.80 m/s )(0.010Ω )

 v = mg ⇒ vterm = 2 2 =
= 0.0245 m/s
 R  term
(0.20 m )2 (1.0 T)2
l B
b) The loop has to fall 10 cm for the top edge to leave the field. This takes
If there were no field, the loop would drop in free fall. In this case,
Assess: It takes the loop much longer to fall in the field. This is an example of magnetic braking.
Problem 49.
Model: Assume the magnetic field is uniform over the loop.
Visualize: As the wire falls, the flux into the page will increase. This will induce a current to
oppose the increase, so the induced current will flow counterclockwise. As this current passes
through the slide wire, it experiences an upward magnetic force. So there is an upward force – a
28
retarding force – on the wire as it falls in the field. As the wire speeds up the retarding force will
become larger until it balances the weight.
Solve: a) & b) The force on the current-carrying slide wire is Fm = IlB. The induced current is
ε 1 dΦ 1 d
B d
Blv
I= =
=
AB =
lx =
R R dt
R dt
R dt
R
Consequently, the retarding magnetic force is
l 2B 2 v  l 2B 2 
 v
Fm =
= 
 R 
R
The important point is that Fm is proportional to the speed v. As the wire begins to fall and its
speed increases, so does the retarding force. Within a very short time, Fm will increase in size to
where it matches the downward weight force w = mg. At that point, there is no net force on the
loop, so it will continue to fall at constant speed.
The condition that the upward magnetic force magnitude equal the weight magnitude is
2
 l2B2 
mgR (0.010 kg)(9.80 m/s )(0.10Ω )

 v = mg ⇒ vterm = 2 2 =
= 0.98 m/s
 R  term
(0.20 m)2(0.50 T )2
l B
Assess: This is a reasonable speed.
Problem 50.
Model: Assume the coil is moved to a location where the magnetic field is zero.
Visualize: The removal of the coil from the field will change the flux and produce an induced
emf and corresponding induced current. The current will charge the capacitor.
Solve: The induced current is
ε coil
N dΦ
R
R dt
But the definition of current is I = dq/dt. Consequently, the charge flow through the coil and onto
the capacitor is given by
∆q N ∆Φ
N
N
dq N dΦ
=
⇒
=
⇒ ∆q =
∆Φ =
(Φ f − Φi )
R dt
∆t
R ∆t
R
R
dt
We are only interested in the total charge that flows due to the change in flux and do not care
about the details of the time dependence. In this case, the flux is changed by physically pulling
Icoil =
29
=
the coil out of the field. Since the coil is oriented for maximum flux the initial flux through the
coil is Φi = π r2B = π (0.0050 m)2(0.0010 T) = 7.85×10–8 Wb. After being pulled from the field,
the final flux is Φf = 0. The charge that flows onto the capacitor is
10 (0 Wb − 7.85 × 10 −8 Wb)
N
−6
∆q = (Φf − Φ i ) =
= 3.93 × 10 C
0.20Ω
R
This charge causes the capacitor to be charged to a potential difference
∆q 3.93 × 10−6 C
∆VC =
=
= 3.93 V
1.0 × 10 −6 F
C
Problem 51.
Model: Assume the field is uniform in the region of the coil.
Visualize: The rotation of the coil in the field will change the flux and produce an induced emf
and a corresponding induced current. The current will charge the capacitor.
Solve: The induced current is
ε coil
N dΦ
R
R dt
But the definition of current is I = dq/dt. Consequently, the charge flow through the coil and onto
the capacitor is given by
∆q N ∆Φ
N
N
dq N dΦ
=
⇒
=
⇒ ∆q = ∆Φ = (Φf − Φ i )
R dt
∆t
R ∆t
R
R
dt
We are only interested in the total charge that flows due to the change in flux and do not care
about the details of the time dependence.
r r In this case, the flux is changed by physically rotating
the coil in the field. The flux is Φ = A ⋅ B = AB cosθ so the change in flux is given by
∆Φ = AB (cosθ f − cosθ i ) = π r 2 B(cosθ f − cosθ i )
Icoil =
=
= π ( 0.020 m )2 (55 × 10 −6 T)(cos30 − cos210) = 1.2 × 10 −7 Wb
Note that the field is 60o from horizontal and the surface normal to the loop is vertical so the
initial angle is 30o and after rotation it is 210o. The charge that flows onto the capacitor is
200(1.2 × 10 −7 Wb)
N
−5
∆q = (Φ f − Φi ) =
= 1.2 × 10 C
R
2.0Ω
This charge causes the capacitor to be charged to a potential difference
∆q 1.2 × 10 −5 C
∆VC =
=
= 12 V
C 1.0 × 10−6 F
30
Problem 52.
Visualize: The battery will cause a current through the inductor and hence stored energy.
Solve: The energy stored in the inductor depends on the inductance and the current. We can find
the current from Ohm's law.
∆V
εbat = 12 V = 2.0 A
I=
=
R
Rind + r ( 4.0 + 2.0)Ω
UL = 12 LI 2 = 12 (100 × 10 −3 H)(2.0 A ) = 0.20 J
2
Problem 53.
Model: Assume we can use the magnetic field of a solenoid.
Visualize: The current through the inductor has energy stored in the magnetic field.
Solve: The energy stored in the inductor depends on the inductance and the current. We know
the energy and the current so we can get the inductance.
−3
2UL 2(1.0 × 10 J)
2
−2
1
UL = 2 LI ⇒ L = 2 =
= 5.0 × 10 H
2
I
(0.20 A )
We know the flux per turn for a given current so we can relate the inductance to the magnetic
field and the flux per turn
 µ NI 
 µ NA 
Φ (1) = BsolA =  0  A =  0  I where Φ (1) is the flux through one turn of the solenoid.
 l 
 l 
 µ NA  L
µ 0 N 2 A  µ0 NA 
=
 N ⇒  0  = sol . So we can substitute in
But the inductance is Lsol =
 l 
 l 
N
l
−2
L 
(5.0 × 10 H)(0.200 A ) = 500 turns
L
(1)
and get Φ =  sol  I ⇒ N = sol
( 1) I =
 N 
Φ
20 × 10−6 Wb/turn
Problem 54.
Model: Assume we can use the magnetic field of a solenoid.
Visualize: The changing current produces an emf across the inductor.
Solve: When the current is changing we know the induced emf so we can get the inductance.
dI
∆V
0.20 V
∆V = − L
⇒ L = dI =
= 20 mH
dt
dt 10.0 A/s
We know the flux per turn for a given current so we can connect the inductance to the magnetic
field and the flux per turn
31
 µ NI 
 µ NA 
Φ (1) = BsolA =  0  A =  0  I where Φ (1) is the flux through one turn of the solenoid.
 l 
 l 
 µ NA  L
µ N 2 A  µ0 NA 
=
 N ⇒  0  = sol . So we can substitute in
But the inductance is Lsol = 0



l
l
l 
N
−3
L 
(20 × 10 H)(0.10 A ) = 400 turns
L
(1)
and get Φ =  sol  I ⇒ N = sol
( 1) I =
 N 
Φ
5.0 × 10−6 Wb/turn
Problem 55.
Model: We will take the wire to be "infinite". Assume the solenoid is long enough that we can
take the field to be constant.
Visualize: The energy density depends on the field strength in a particular region of space.
Solve: a) We can use the standard "infinite" wire result to find the energy density.
µ0 I
(4π × 10−7 T m/A )(1.0 A ) = 4.0× 10 −4 T
B=
=
2π rwire
2π (0.00050 m )
4.0× 10 −4 T)
(
1 2
uB =
B =
= 6.37× 10−2 J/m 3
−7
2 µ0
2(4π × 10 T m/A )
b) We can find the magnetic field of the solenoid. The number of turns and length of the
solenoid are not known. However, the length can be related to the diameter of the winding wire,
l = Ndwire , which we do know. Using the field we determine the energy density.
−7
µ0 NI µ0 NI (4π × 10 T m/A )(1.0 A )
−3
B=
=
=
= 1.26× 10 T
l
Ndwire
(0.0010 m )
2
(1.26× 10−3 T) = 0.628 J/m 3
1 2
uB =
B =
−7
2 µ0
2(4π × 10 T m/A )
2
Assess: Note that the energy density is larger in the solenoid for the same current.
Problem 56.
Model: We will take the straight wire to be "infinite".
Visualize: The energy density depends on the field strength in a particular region of space.
Solve: a) We can use the result for the field at the center of a current loop and then find the
energy density.
32
−7
µ0 I (4π × 10 T m/A )(1.0 A )
B=
=
= 3.14× 10 −5 T
2rloop
2(0.020 m )
(3.14× 10 T ) = 3.93× 10−4 J/m 3
1 2
B =
uB =
2 µ0
2(4π × 10 −7 T m/A )
−5
2
b) We use the standard result for a long straight wire.
2
1 2
1  µ0 I 
µ I


B = 0 , uB =
B =
2 µ0
2µ 0  2π r 
2π r
⇒
8π r uB
=
µ0
2 2
I=
8π 2 (0.020 m ) (3.93 × 10−4 J/m 3 )
2
(4π × 10
−7
T m/A )
= 3.14 A
Problem 57.
Model: Take the magnetic field to be constant throughout the atmosphere.
Visualize: The energy density depends on the field strength in a particular region of space. The
total energy is the energy density times the volume.
Solve: a) We are given enough to find the energy density so we need to determine the volume.
The atmosphere is a spherical shell between the surface of the earth at Re and the top of the
atmosphere Re+h where h is the thickness of the atmosphere.
2
50× 10 −6 T )
(
1 2
uB =
B =
= 9.95 × 10−4 J/m3
−7
2 µ0
2(4π × 10 T m/A )
[
] [
]
Vatm = 43 π (Re + h) − Re = 43 π (6.37 × 10 m + 20 × 10 m ) − (6.37 × 10 m ) = 1.02 × 10 m
3
3
Utot = uBVatm = (9.95 × 10
−4
J/m
3
6
)(1.02 × 10
3
19
m
3
) = 1.0 × 10
3
16
6
3
19
3
J
b) Find the ratio Umag / Utot = 1.0 × 101 6 J/4.0 × 101 8 J = 0.25 × 10 −2 , so its 0.25%.
Assess: Not enough to do much good.
Problem 58.
Model: Assume the solenoid is long enough that we can approximate the field as constant inside
and zero outside.
Visualize: The energy density depends on the field strength in a particular region of space.
Solve: a) We can use the given field strength to find the energy density and then determine the
volume of a cylinder.
33
(5.0T)
1 2
B =
= 9.95 × 106 J/m3
−7
2 µ0
2(4π × 10 T m/A )
2
uB =
Vsol = πrsol2l = π (0.20 m ) 1.0 m = 0.126 m 3
2
Utot = uBVsol = (9.95 × 10 6 J/m 3 )(0.126 m 3) = 1.25 × 106 J
b) We use the magnetic field of the solenoid.
µ NI
(1.0 m )(5.0 T )
lB
4
B= 0
⇒ N=
=
= 4.0× 10 turns
−7
µ0 I (4π × 10 T m/A )(100 A )
l
Assess: This is a reasonable number for a meter long solenoid.
Problem 59.
Model: Assume the field is constant over the cross section of the patient.
Visualize: The changing field results in a changing flux and an induced emf.
Solve: Assume the surface normal is parallel to the field to determine the largest emf. From
dΦ
d
dB
dB
∆B
=
AB = A
=A
=A
Faraday's law we have ε =
. Let's determine the time
dt
dt
dt
dt
∆t
interval acceptable for the change.
2
∆B (0.060 m )(5.0 − 0) T
∆t = A
=
= 3.0 s
ε
0.10 V
Assess: This is a reasonable amount of time in which to change the field.
Problem 60.
Model: Assume the magnetic field is uniform over the loop.
Visualize: The motion of the eye will change the orientation of the loop relative to the fixed
field direction resulting in an induced emf.
Solve: We are just interested in the emf due to the motion of the eye so we can ignore the details
of the time dependence of the change. From Faraday's law
2
ε coil = N dΦ = N ∆Φ = NAB ∆ cosθ = Nπr Bcos θ f − cosθ i
dt
∆t
∆t
∆t
20π (3 × 10 −3 m ) (1.0 T) cos5 − cos0
=
= 1.1× 10 −5 V
0.20 s
2
Assess: This is a reasonable emf to measure although you might need some amplification.
34
Problem 61.
Model: Assume we can ignore the sharp corners when the current changes abruptly.
Visualize: The changing current produces a changing flux and an induced emf and
corresponding potential difference.
Solve: Break the current into time intervals over which the current is changing linearly or not at
all. For the intervals 1 to 2 ms and 4 to 5 ms the current does not change so the potential
difference is zero. On the interval 0 to 1 ms the current goes from 0 to 1 A so the potential
difference is
dI
∆I
(1 − 0)A
∆V = − L = − L
⇒ ∆V = − (10 × 10−3 H)
= −10 V
dt
∆t
(1 − 0) × 10 −3 s
Similar calculations give: 2 to 4 ms, -5 V; 5 to 6 ms, +20 V. See the figure.
Assess: The potential difference is proportional to the negative slope of the current vs. time.
Problem 62.
Model: Assume we can ignore the sharp corners when the current changes abruptly.
Visualize: The changing current produces a changing flux and an induced emf and
corresponding potential difference.
Solve: Break the current into time intervals over which the current is changing linearly or not at
all. For the intervals 2 to 3 ms and 5 to 6 ms the current does not change so the potential
difference is zero. On the interval 0 to 2 ms the current goes from 1 to -1 A so the potential
difference is
dI
∆I
(−1 − 1)A
∆V = − L = − L
⇒ ∆V = − (10 × 10−3 H)
= 10 V
dt
∆t
(2 − 0) × 10 −3 s
Similar calculations give: 3 to 4 ms, -10 V; 4 to 5 ms, +20 V. See the figure.
35
Assess: The potential difference is proportional to the negative slope of the current vs. time.
Problem 63.
Model: Assume we can ignore the edges when the potential difference changes abruptly.
Visualize: The existence of a potential difference indicates that there must be a changing
current. If the potential difference is zero that means the current is constant.
Solve: Break the potential difference into time intervals and determine the corresponding rate of
change of the current each time interval. Knowing the starting current for an interval we can
determine the current at the end of the interval by finding the change. For the intervals 10 to 20
ms and 30 to 40 ms the potential difference is zero so the current remains the same as at the start
of the interval. On the interval 0 to 10 ms the potential difference is 1 V so the rate of change of
the current is
(1 V )(10 - 0) × 10 −3 s
dI
∆I
− ∆V∆t
∆V = − L = − L
⇒ ∆I0→1 0 =
=−
= −0.20 A
50 × 10 −3 H
dt
∆t
L
To find the current at 10 ms we need the current at the start of the interval and the change.
I(10) = I(0) + ∆I0→1 0 = 0.20 A + (−0.20 A ) = 0
A constant potential difference implies that the current is changing linearly. So the current starts
at 0.2 A and goes linearly to 0 A over the first 10 ms. Similar calculations give: 10 to 20 ms,
remain at 0 A; 20 to 30 ms, go linearly from 0 to -0.4 A; from 30 to 40 ms, remain at -0.4 A.
36
Problem 64.
Model: Assume we can ignore the edges when the potential difference changes abruptly.
Visualize: The existence of a potential difference indicates that there must be a changing
current. If the potential difference is zero that means the current is constant.
Solve: Break the potential difference into time intervals and determine the corresponding rate of
change of the current each time interval. Knowing the starting current for an interval we can
determine the current at the end of the interval by finding the change. For the intervals 0 to 10
ms and 30 to 40 ms the potential difference is zero so the current remains the same as at the start
of the interval. On the interval 10 to 20 ms the potential difference is 1 V so the rate of change
of the current is
dI
∆I
−∆V∆t
(1 V )(20 -10 ) × 10−3 s
∆V = − L = − L
⇒ ∆I1 0→ 2 0 =
=−
= −0.10 A
dt
∆t
L
100 × 10 −3 H
To find the current at 20 ms we need the current at the start of the interval and the change.
I(20) = I(10) + ∆I1 0→ 2 0 = 0.10 A + (−0.10 A ) = 0
A constant potential difference implies that the current is changing linearly. So the current starts
at 0.10 A and goes linearly to 0 A from 10 to 20 ms. Similar calculations give: 0 to 10 ms,
remain at starting 0.10 A; 20 to 30 ms, go linearly from 0 to 0.10 A; 30 to 40 ms, remain at 0.10
A.
Problem 65.
Visualize: The current through the inductor is changing with time which leads to a changing
potential difference across the inductor.
Solve: a) To find the potential difference we differentiate the current.
 −1
 LI 
dI
d
∆VL = − L = −L (I0 e − t / τ ) = − L  I0 e −t / τ =  0  e −t / τ
τ 
 τ 
dt
dt
b) For the potential difference at a particular time we just plug in and evaluate. Here we do it
for t = 1 ms.
37
 20 × 10 −3 H 50 × 10 −3 A 
 LI 
(
)(
) e −t / 1.0×1 0−3 s = (1.0 V )e− t / 1.0 ×1 0−3
∆VL =  0  e −t / τ = 
−3

 τ 
1.0 × 10 s


∆VL (1 ms) = (1.0 V )e
−1.0×1 0−3 s /1.0×1 0−3 s
s
= (1.0 V )e = 0.37 V
−1
Similar calculations give: 0 s, 1.0 V; 1 ms, 0.37 V; 2 ms, 0.13 V; 3 ms, 0.05 V.
c) See the graph below.
Problem 66.
Visualize: The potential difference across the inductor depends on the rate of change of the
current with respect to time.
Solve: a) We can find the potential difference by taking a derivative
dI
d
∆VL = − L = −L I 0 sin ωt = − Lω I0 cosωt = −2π fLI0 cos ωt
dt
dt
b) The cosine function oscillates between +1 and -1 so we ignore it and the negative sign.
∆VL ,max
0.20 V
-3
∆VL ,max = 2πfLI0 ⇒ I0 =
=
= 1.27 × 10 A
2π (500 × 10 3 Hz)(50 × 10 -6 H)
2πfL
Problem 67.
Model: Assume any resistance is negligible.
Visualize: The potential difference across the inductor and capacitor oscillate so that they are
out of phase by one-quarter cycle.
Solve: a) The current is given by I(t) = I0 cosωt = (0.50 A )cosωt . Looking at Fig. 33-45 we
see that the capacitor is fully charged one-quarter cycle after the current is a maximum (or
minimum) so the time needed is one quarter cycle.
38
ω=
T=
1
=
LC
1
(20 × 10
−3
H)(8.0 × 10
−6
F)
= 2.5 × 103 rad/s
2π
2π
=
= 2.5 × 10 −3 s = 2.5 ms ⇒
3
ω
2.5 × 10 rad/s
∆t = 14 T = 0.625 ms
b) The charge and the current are related by a derivative so I0 = ωQ0 . Now we get the potential
across the capacitor
Q
I
0.50 A
∆VC = 0 = 0 =
= 25 V
3
C ωC (2.5 × 10 rad/s )(8.0 × 10-6 F)
Problem 68.
Model: Assume any resistance is negligible.
Visualize: The potential difference across the inductor and capacitor oscillate. The period of
oscillation depends on the resistance and capacitance and the potential difference across the
capacitor depends on the charge.
Solve: The current is given by I(t) = I0 cosωt = (0.60 A )cosωt . We can relate the extrema of
the current and the capacitor potential difference. Using ω = 1 LC and Q = C∆VC we find
1
C
I0 = ωQ0 = ω C∆VC,max =
C∆VC,max =
∆VC, max
LC
L
I02
(0.60 A )
−3
L=
H)= 1.0 × 10−6 F = 1.0 µF
2 (10 × 10
2
∆VC,max
(60 V )
2
C=
Assess: This is a reasonable capacitance.
Problem 69.
Model: Assume any resistance is negligible.
Visualize: The energy in the capacitor and the energy in the inductor oscillate as the charge and
the current change.
Solve: We know the capacitance and the inductance. The charge on the capacitor is given by
Q(t) = Q0 sin ωt . To determine the energies we will also need the current
d
d
I(t) = Q(t) = (Q0 sin ωt ) = ωQ0 cosω t .
dt
dt
Now let's set the energies equal UL = U C ,
2
2
2
2
2
2
1
1 Q
LI
=
⇒ LC(ωQ0 cos ωt ) = (Q0 sin ωt ) ⇒ (cosω t ) = (sin ωt )
2
2
C
39
1
π
. This will be satisfied for tan ωt = 1 , which occurs when ωt = . Let's find the
LC
4
π
π
current at this time Q(ωt = 4 ) = Q0 sin 4 = 0.707Q0 .
using ω =
Problem 70.
Model: We will design a solenoid inductor.
Visualize: The oscillation frequency depends on the capacitance and the inductance. The
inductance will depend on the number of turns, the length and the cross-sectional area.
Solve: First we'll find the inductance necessary to give the proper frequency, then we can design
our inductor. For an LC circuit
1
1
1
−2
ω=
= 2πf ⇒ L =
=
H
2 2
2
2
2 = 2.53 × 10
LC
4π f C 4π (1.0 Hz) (1.0 F )
2
2
Now the inductance of a solenoid is Lsol = µ 0 N A / l . The area is A = π r . The length depends
on the number of turns, l = Nd , where d is the diameter of the wire used. So we find that
dL
µ N 2 A µ0 Nπr 2
(0.25 × 10 −3 m)(2.53 × 10 −2 H) = 4.0 × 103 turns
L= 0
=
⇒ N=
=
d
µ0 πr 2 (4π × 10 −7 T m/A )π (0.02 m)2
Nd
−3
Lots of turns, if we have one layer the length will be l = Nd = 4000(0.25 × 10 m) = 1.0 m
which is pretty long.
Assess: We would probably want to have several layers and a shorter solenoid.
Problem 71.
Model: Assume negligible resistance in the LC part of the circuit.
Visualize: With theswitch is position 1 for a long time the capacitor is fully charged. After
moving the switch to position 2 we will have oscillations in the LC part of the circuit.
Solve: a) After a long time the potential across the capacitor will be that of the battery and
Q0 = C∆Vbatt . When the switch is moved the charge will move giving a current
1
C
I0 = ωQ0 = ω C∆Vbatt =
C∆Vbatt =
∆Vbatt
LC
L
I0 =
(2.0 × 10
(50 × 10
−6
−3
F)
H)
(12 V ) = 7.6 × 10 −2 A = 76 mA
b) The current will be maximum one-quarter cycle after the maximum charge, so we find the
2π
= 2π LC = 2π (50 × 10 −3 H)(2.0 × 10 −6 F) = 2.0 ms , so we get
period T =
ω
∆t = 14 T = 0.50 ms .
Problem 72.
40
Model: Assume negligible resistance in the circuit.
Visualize: The maximum voltage on the right capacitor will occur when all of the energy from
the left capacitor is transferred to the right one.
Solve: a) The energies are related by
C1
300
∆VC,1 =
(100 V ) = 50 V .
C2
1200
b) After closing S1 , wait one-quarter period of the left LC combination so the left capacitor is
fully discharged and the current is maximum in the inductor, then open S1 and close S2. The
current will the flow to the right capacitor. Let's find the period of the left LC combination,
2π
T=
= 2π LC = 2π (5.3 H)(300 × 10 −6 F) = 0.25 s . The time is ∆t = 14 T = 0.0625 s .
ω
1
2
C1 ∆VC ,1 = 12 C2∆VC, 2 ⇒ ∆VC,2 =
2
2
Problem 73.
Visualize: The two parts of the circuit are in parallel directly across the battery. When the
switch is closed there is current in the left resistor, it's just a battery and a resistor. The behavior
in the inductor branch changes with time because the potential difference depends on the rate of
change of the current.
Solve: a) The current through the battery is the total of the current through the two branches.
Just after the switch is closed, because the current is changing rapidly, the potential difference
across the inductor (back emf) is large and will oppose any current in the inductor branch. The
inductor looks like an open circuit, all the current is through the left resistor.
∆V
10 V
Ibatt = Ileft + Iright =
+0=
= 1.0 A
R
10 Ω
b) After a long time there is still the same current in the left resistor. The rate of change of the
current in the inductor is small, so the potential difference across the inductor is essentially zero,
and the inductor behaves like a short circuit.
∆V
∆V
10 V
Ibatt = Ileft + Iright =
+
=2
= 2.0 A
Rleft Rright
10 Ω
Problem 74.
Visualize: The two branches behave differently. When the switch is closed there is current in
the left resistor. The behavior in the inductor branch changes with time because the potential
difference across the inductor depends on the time rate of change of the current.
Solve: a) Just after the switch is closed, because the current is changing rapidly, the potential
difference across the inductor (back emf) is large and will oppose any current through the
inductor. The inductor appears like an open circuit and the current is through the two resistors.
This is just a battery with two resistors in series so they have the same current.
41
I2 0 =
∆V
30 V
=
= 3.0 A
Rtop + Rmiddle (10 + 20) Ω
b) After the switch has been closed a long time the time rate of change of the current through the
inductor is small, so the potential difference across the inductor is essentially zero, and the
inductor behaves like a short circuit. At this point the current will essentially all flow through
the top resistor and the inductor bypassing the 20 Ω resistor.
10 V
I1 0 =
= 1.0 A, I2 0 = 0 A
10 Ω
c) Just after the switched is reopened the top resistor will have no current through it. The
inductor will develop a back emf and initially try to keep the current at 1.0 A which now goes
through the 20 Ω resistor, I2 0 = 1.0 A .
Problem 75.
Visualize: The inductor will initially oppose the current by developing a back emf so the current
will be small. After a while the time rate of change of the current will become less, the back emf
will decrease, and the inductor acts like a wire so the current is maximum.
Solve: a) After a long time the current is not changing so the inductor has no potential difference
across it and we have essentially a battery with a resistor.
∆Vbatt
∆Vbatt + ∆VR = ∆Vbatt − IR = 0 ⇒ I = I0 =
R
b) Now let's look at the general case and apply Kirchoff's loop rule to the circuit.
dI
∆Vbatt + ∆VR + ∆VL = ∆Vbatt − IR − L = 0
dt
L dI
dI R
= I0 − I ⇒
= (I − I )
R dt
dt L 0
where we used the definition of I0 from part a. We also know that because of the back emf at the
start I = 0 for t = 0. After a long time when the current is established , I = I0 for t → ∞ . This
looks almost like the differential equation for the exponential function except for the constant I0
dz
dI
= − , now we can rewrite our
on the right side. Let's define a new variable z = I0 − I so
dt
dt
dz
R
−t /( L / R )
= − z ⇒ z = Ae
differential equation
. We now determine the constant A from the
dt
L
0
initial conditions, for t = 0 we have I = 0 so z(0) = Ae = A = I0 . Finally we find the current as a
function of time
z = I0 e
−t /( L / R)
= I0 − I
42
⇒ I(t) = I0 (1 − e
−t /( L / R )
)
Assess: The current is zero at the start and approaches the steady final value. The behavior is
similar to the charging of a capacitor.
Problem 76.
Visualize: The area within in the loop is changing so the flux will change. This will produce
and induced emf and corresponding current.
Solve: a) The field is out of the page, so the flux is increasing outward as the loop expands.
According to Lenz’s law, the induced current must try to prevent the increase of outward flux. It
does so by generating a field into the page. This requires a clockwise current flow.
r
r
b) Take A parallel to B , so Φ = AB and ε = |dΦ/dt| = B(dA/dt). Here the field is constant but the
loop area is changing. Let the length of each edge of the loop be x at time t. This length is
increasing linearly with time as the corner of the loop moves outward with speed v = 10 m/s:
x = x0 + vxt = 0 + [(10 m/s)cos45°]t = 7.07t m.
The loop’s area at time t is
A = x2 = (7.07t m)2 = 50t2 m 2
dA
⇒
= 100 t m 2 / s.
dt
Consequently, the induced emf at time t is
ε = B dA = (0.1 T)(100t m 2 / s) = 10t V.
dt
The induced current is I = ε/R, where R is the resistance of the loop. The wire has resistance 0.01
Ω per meter, so R = 0.01 l where l = 4x is the perimeter of the square at time t. That is,
R = 4(0.01 Ω/m)(7.07t m) = 0.282t Ω.
43
Thus the induced current is
I=
ε
R
=
10tV
= 35.5 A,
0.282t Ω
which is independent of t.
c) At t = 0.1 s, ε = 1.0 V and I = 35.5 A.
Assess: The induced emf depends on time, but so does the resistance so the induced current is
constant.
Problem 77.
Model: Assume the straight wire is very long so we can use the "infinite" wire result.
Visualize: As the short wire moves in the field of the long wire the charges will experience a
magnetic force. They will move inside the wire until the electric field created by the charge
separation has a corresponding force that balances the magnetic force. The potential difference
between the ends of the wire is due to this electric field.
Solve: To find the potential difference between the ends of the short wire we need to know the
electric field along it. The magnetic field of the long wire is into the page near the short wire so
the magnetic force on positive charges will be to the left. Positive charges will move to the left,
leaving negatives on the right. This creates an electric field and corresponding force to the right.
Look at the charges at a distance x from the long wire. In equilibrium, the magnetic and electric
µ I
forces balance, qvB = qE ⇒ E = vB = v 0 using the magnetic field for an "infinite" wire.
2π x
Find the potential difference by integrating the electric field from the negative end to the
positive. Calculating the potential difference,
r d+ l  vµ I 
 vµ I  d +l dx  vµ0 I   d + l 
d r
 ln 
∆V = − ∫d+El ⋅dl = ∫d  0  xˆ ⋅ dx xˆ =  0  ∫d

= 
 2π x 
 2π 
x  2π   d 
r
where we have used dl = dx xˆ .
Assess: Seems reasonable, the potential difference grows with current (stronger magnetic field)
and velocity (larger motional emf).
44
Problem 78.
Model: Assume that the sides remain straight during the collapse. Also, take the magnetic field
to be constant in the region of interest.
Visualize: As the loop collapses the area will change and there will be a corresponding change
in flux and therefore an induced emf and current.
Solve: a) For simplicity we will focus on the upper-right triangle. The width of the upper-right
triangle is given by x = x0 + vt , with x0 = b 2 = (0.10 m ) 2 = 0.0707 m and v = 0.293 m/s.
The height y is related to the width x since it's part of a triangle, y = b 2 − x 2 . The loop will be a
straight line when y = 0 and the width is b, xcollapse = b = x 0 + vt collapse . We solve for the time
b − x0 (0.10 − 0.0707) m
t collapse =
=
= 0.100 s
v
0.293 m/s
b) The induced current is due to the changing flux as a result of the changing area
ε 1 dΦ = B dA
I= =
R dt
R R dt
The total area at any time is given by A = 4 × (area of triangle) = 4 × 12 xy = 2x b 2 − x 2 where x
and y are as given above and depend on time. Using the chain rule we find
 2
2 
 b2 − 2x 2 
b − 2(x 0 + vt ) 
d
dA dA dx dA

2
2
 = 2v
=
=
v=v
2x b − x = 2v  2
 b 2 − (x + vt )2 
dx dt dx
dx
dt
 b − x2 


0
 2

2 
2 
2
2vB  b − 2(x0 + vt )  2(0.293 m/s)(0.50 T )  ( 0.10 m ) − 2(0.0707 m + (0.293 m/s)t ) 
I=

=

2
2

0.10Ω
R  b2 − (x + vt )2 


0
 (0.10 m ) − (0.0707 m + (0.293 m/s)t ) 
(
)

2 
2
 ( 0.10 m ) − 2(0.0707 + (0.293)t ) 
= 2.93 
A
 (0.10 m )2 − (0.0707 + (0.293)t )2 


for t in seconds, and we have taken the absolute value.
c) We plug in a few times and get the currents in the table and the corresponding plot.
45
t (s)
I (A)
0.000
0.020
0.040
0.060
0.080
0.090
0.100
0.000
0.078
0.186
0.348
0.671
1.078
diverges
Problem 79.
Model: Assume the field changes abruptly at the boundary and is constant.
Visualize: The loop moving through the field will have a changing flux. This will produce an
induced emf and corresponding current. The induced current will oppose the change in flux and
result in a retarding force on the loop.
Solve: a) The loop moving into the outward field will have an increasing outward flux. The
induced current will oppose this change and create a field that is into the page. This requires that
the induced current flow clockwise. The current-carrying wires of the loop will experience a
force. For a clockwise current the force on the leading edge is to the left, retarding the motion.
The forces on the top and bottom edges cancel and the trailing edge experiences no force since
the field is zero there.
46
ε
1 dΦ B dA Bl dx Blv
=
=
=
. With this
R dt
R dt
R
R R dt
 B 2l 2 
 Blv 
 v remembering
 lB = 
current we can find the force on the leading edge, F = IlB = 
 R 
 R 
We can find the induced current I =
=
that it opposes the motion. Now we can use Newton's second law
 B2 l 2 
 B 2l 2 
dv
dv
 v ⇒
 v .
ma = m
= − 
= − 
 R 
 mR 
dt
dt
We recognize this differential equation as having the exponential function as a solution, so
l 2 B2
v = v 0e − bt with b =
and the initial velocity is v0.
mR
b) We plot the velocity knowing that b = 100 s using the numbers given.
Assess: This exponential decay of the velocity is called eddy-current braking.
Problem 80.
Model: Assume the field changes uniformly when turned on and that the loop is initially at rest.
Visualize: The changing magnetic field produces a changing flux that will generate an induced
emf and current. The current-carrying wire will experience a force that opposes the change.
47
Solve: The increasing field strength produces an increasing outward flux. The induced current
will oppose this change and create a field that is into the page. This requires that the induced
current flow clockwise. The current-carrying wires of the loop will experience a force.
r
r
r r
a) Take A parallel to B so the flux is Φ = A ⋅ B = AB . Because the field strength is changing
dΦ 1  dB dA 
= A
+
B . We will assume
and the loop is moving the emf has two terms, ε =
dt
R  dt
dt 
that the loop does not move much and ignore the dA/dt term (see below). With A = l(l / 2) ,
ε 1 dB = l 2 ∆B = (0.080 m )2 (100 A/s) = 32 A
I= = A
R R dt 2R ∆t
2(0.010Ω )
b) For a clockwise current the force on the leading edge is to the left, so it is away from the
magnetic field. The forces on the top and bottom edges cancel and the trailing edge experiences
no force since the field is zero there. So we find
F = IlB = (32 A )(0.080 m )(0.50 T ) = 1.28 N .
c) We use Newton's second law, a = F / m = (1.28 N) / (0.010 kg) = 128 m/s . If acceleration is
2
2
2
−3
constant ∆s = 12 a(∆t ) = 12 (128 m/s )(0.010 s) = 6.4 × 10 m = 0.64 cm
2
d) The distance travelled is about a factor of ten smaller than the size of the loop, so ignoring the
motion is reasonable.
e) To find the speed after being "kicked" we calculate the impulse. We can use the average force
found in the previous part.
F ∆t (128 N)(0.010 s)
Fave ∆t = ∆p = mvf − mvi ⇒ vf = ave =
= 1.28 m/s
m
(0.010 kg)
Assess: This shows that the loop would be shot out of the field region.
Problem 81.
Model: Assume the conductor is long enough to use the "infinite" wire magnetic field.
Visualize: The current in the inner conductor creates a field that circles it in the space between
the inner and outer conductor and there is corresponding flux there.
48
r
Solve: a) We
of length l as shown. Take A
r r
r will look at a rectangle between the conductors
parallel to B so the flux through the strip is dΦ = B ⋅ dA = Bldr . We take a strip all at the same
radius because the field strength depends only on the distance from the inner conductor. Let's
find the total flux
r2  µ I 
µ Il r2 dr µ0 Il  r2 
Φ = ∫ dΦ = ∫r  0  l dr = 0 ∫r
=
ln 
1  2πr 
2π 1 r
2π  r1 
where we have used the field of an infinite wire. We want the inductance per unit length so we
take the inductance and divide by the length.
r 
Φ
Φ µ
L=
⇒ L˜ = = 0 ln  2 
I
Il 2π  r1 
b) Now we just plug in numbers
 r  (4π × 10 −7 T m/A )  3.0 
Φ µ
−7
 = 3.6 × 10 H/m = 0.36 µH/m
ln 
L˜ = = 0 ln  2  =
 0.50 
Il 2π  r1 
2π
49