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Kapittel 34
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34.1. Visualize:
To develop a motional emf the magnetic field needs to be perpendicular to both, so let’s say its direction is into the
page.
Solve: This is a straightforward use of Equation 34.3. We have
v
E
1.0 V

 2.0  104 m/s
lB 1.0 m   5.0  105 T 
Assess: This is an unreasonable speed for a car. It’s unlikely you’ll ever develop a volt.
34.4. Model: Assume the field changes abruptly at the boundary between the two sections.
Visualize: Please refer to Figure Ex34.4. The directions of the fields are opposite, so some flux is positive and some
negative. The total flux is the sum of the flux in the two regions.
Solve: The field is constant in each region so we will use Equation 34.10. Take A to be into the page. Then, it is
parallel to the field in the left region so the flux is positive, and it is opposite to the field in the right region so the flux
is negative. The total flux is
  A  B  AL BL cos L  AR BR cos R
  0.20 m   2.0 T    0.20 m  1.0 T   0.040 Wb
2
2
Assess: The flux is positive because the areas are equal and the stronger field is parallel to the normal of the surface.
34.5. Model: Consider the solenoid to be long so the field is constant inside and zero outside.
Visualize: Please refer to Figure Ex34.5. The field of a solenoid is along the axis. The flux through the loop is only
nonzero inside the solenoid. Since the loop completely surrounds the solenoid, the total flux through the loop will be
the same in both the perpendicular and tilted cases.
Solve: The field is constant inside the solenoid so we will use Equation 34.10. Take A to be in the same direction as
the field. The magnetic flux is
  Aloop  Bloop  Asol  Bsol   rsol2 Bsol cos    0.010 m   0.20 T   6.3  105 Wb
2
When the loop is tilted the component of B in the direction of A is less, but the effective area of the loop surface
through which the magnetic field lines cross is increased by the same factor.
34.6. 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
because 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 to be into the page so it is in the same direction as the
magnetic field. The total flux is


loop
B  dA 

inside
rectangle
B  dA 

outside
rectangle
B  dA 

inside
rectangle
BdA cos  BAinside
 Bab
rectangle
Assess: Only the region with nonzero magnetic field inside the loop contributes to the flux.
34.7. Model: Assume the field strength is uniform over the loop.
Visualize: Please refer to Figure Ex34.7. According to Lenz’s 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.
34.8. Model: Assume that the magnet is a bar magnet with field lines pointing away from the north 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’s 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 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 at the bottom of the solenoid. So, the current will be right to left in 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 signs 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.
34.9. Visualize: Please refer to Figure Ex34.9. The changing current in the solenoid produces a changing flux in
the loop. 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. As the current in the solenoid increases there is more field and more 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
left. This requires a counterclockwise current.
34.11. Model: Assume the field is uniform.
Visualize: Please refer to Figure Ex34.11. If the changing field produces a changing flux in the loop there will be a
corresponding induced emf and current.
Solve: (a) The induced emf is E= d /dt and the induced current is I  E=/R. The field B is changing, but the area A
is not. Take A to be out of the page and parallel to B , so   AB. Thus,
E A
dB
dB
2
  r2
   0.050 m   0.50 T/s   3.9 103 V  3.9 mV
dt
dt
I
E 3.93  103 V

 2.0  102 A  20 mA
R
0.2 
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), E=
 A(dB/dt )  3.9 mV and I  20 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.
(c) Now A (left or right) is perpendicular to B and so A  B  0 Wb. That is, the field does not penetrate the plane of
the loop. If   0 Wb, then E  |d/dt|  0 V/m and I  0 A. There is no induced current.
Assess: Note that the induced field opposes the change.
34.12. Model: Assume the field is uniform.
Visualize: Please refer to Figure Ex34.12. The motion of the loop changes the flux through it. This results in an
induced emf and current.
Solve: The induced emf is E  |d/dt| and the induced current is I /R. The area A is changing, but the field B is
not. Take A as being out of the page and parallel to B , so   AB and d/dt  B(dA/dt). The flux is through that
portion of the loop where there is a field, that is, A  lx. The emf and current are
EB
d  lx 
dA
B
 Blv   0.20 T  0.050 m  50 m/s   0.50 V
dt
dt
I
E 0.50 V

 5.0 A
R 0.10 
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 flows clockwise.
Assess: This seems reasonable since there is rapid motion of the loop.
34.13. 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 the coil is
EN


d A B
d
dB
dB
2  0.10 T 
N
 NA
 N r 2
 103   0.010 m  
  3.1 V
3
dt
dt
dt
dt
 10 10 s 
where we’ve used the fact that B is parallel to A .
Assess: This seems to be a reasonable emf as there are many turns.
34.15. Se en alternative løsning nedenfor
Model: A changing magnetic field creates an electric field.
Visualize: Please refer to Figure Ex34.15 in your textbook.
Solve:
The magnetic field produced by a current I in a solenoid is
Bsolenoid 
0 NI
L
The magnitude of the induced electric field inside a solenoid is given by Equation 34.26. The direction can be found
using Lenz’s law, but the problem only asks for field strength, which is positive.
Only the current is changing, thus
 4 10
E
In the interval t  0 s to t  0.1 s,
7

T m/A  0.010 m  400  dI
2  0.20 m 
dt
V s  dI

 1.26 105
A m  dt

dI
5A

 50 A/s
dt 0.1 s
In the interval t  0.1 s to 0.2 s, the change in the current is zero. In the interval t  0.2 s to 0.4 s, dI dt  25 A/s.
Thus in the interval t  0 s to t  0.1 s, E  6.3  10–4 V/m. In the interval t  0.1 s to 0.2 s, E is zero. In the interval t  0.2
s to 0.4 s, E  3.1  10–4 V/m. The E-versus-t graph is shown earlier, in the Visualize step.
34.15. Alternative løsning:
Det er denne fluksen som skal inn i Faradays lov!
Spole, Nsp = 400
l = 20 cm
B  0 n i  0
l
2
 m  AB   r B

B
Tenkt innerspole, N = 1
r
N sp
0 r 2 N sp
l
i
i
2
d  m 0 r N sp di


dt
l
dt
 rN di

E
 0 sp
2 r
2l dt
di
E  1.256 103
dt
Resten er som løsningen ovenfor.
Hva er E-feltets retning?
I den tenkte sløyfa vil det oppstå en indusert strøm. Feltet har samme retningen som strømmen.
Når spolestrømmen øker vil magnetfeltet i spolen øke. Den induserte strømmen prøver å hindre denne
økningen og lager et motfelt (ut av papirplanet). Stømretningen (og dermed retningen til det induserte
elektriske feltet) blir derfor mot klokka (høyrehåndsregel).
Retningen til det induserte elektriske feltet snur når strømmen begynner å avta.
34.16. Model: A changing magnetic field creates an electric field.
Visualize:
Solve: (a) Apply Equation 34.26. For a point on the axis, r  0 m, so E  0 V/m.
(b) For a point 2.0 cm from the axis,
E
r dB  0.020 m 

  4.0 T/s   0.040 V/m
2 dt 
2

34.17. Model: Assume the magnetic field inside the circular region is uniform. The region looks exactly like a
cross section of a solenoid of radius 2.0 cm. The proton accelerates due to the force of the induced electric field.
Visualize: Please refer to Figure Ex34.17.
Solve: Equation 34.26 gives the strength of the induced electric field inside a solenoid with changing magnetic field.
r dB
The magnitude at each point is thus E 
, where r indicates the radial distance of each point to the center of the
2 dt
circular region.
The direction of E can be determined with Lenz’s Law. As B decreases, a clockwise electric field is induced that
would create a current that opposes the decrease if a conducting loop were present.
Using Newton’s second law,
F  eE  ma  a 
eE er dB

m 2m dt
The sign indicates the acceleration is in the same direction as the induced electric field. At point a the acceleration is
a


 1.6  1019 C 1.0  102 m

2 1.6  10

27
kg

  0.10 T/s   a 

 4.8 10
4
m/s 2 , up



At point b, a  0 m/s2. At point c, a  4.8  10 4 m/s2 , down . At point d, a  9.6  104 m/s2 , down .
34.18. Model: Assume that the current drops uniformly.
Visualize:
Assume that the current is flowing left to right as in the figure. The changing current produces a changing flux in the
inductor and a corresponding emf is developed across the inductor.
Solve: For the inductor the potential difference is
V   L
dI
I
 50 A  150 A  10
 L
  10  103 H 
dt
t
10.0  106 s
3
 100 V
As the current decreases, so does the flux. 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. This induced current
carries positive charge carriers to the right, so the potential will increase in the direction of the current.
34.19. 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:
V   L
dI
I
I
3.0 A  1.0 A
 L
 t  L
  200  103 H 
 1.0 ms
dt
t
Vmax
400 V
Assess: If we change the current in any shorter time the potential difference will exceed the limit.
34.28. 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: (a)
(b) Since the field is perpendicular to the plane of the coil, A is parallel to B and   AB . The emf is
E(t ) 
d
dB
 NA
 20 (0.025 m) 2  0.020  0.020t  T/s  7.9  104 (1  t ) V
dt
dt
 I (t ) 
E(t ) 7.9  104 (1  t ) V

 1.6  103 (1  t ) A
R
0.50 
(c) Using the expression for I(t) in part (b),
I  5 s   1.6 103 1  5 A  9.6 103 A
I 10 s   1.7 102 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.
34.29. 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 loop, and a corresponding
induced emf and current.
Solve: (a) Since the field is perpendicular to the plane of the loop, A is parallel to B and   AB . The emf is
E
d
dB
E
A
 (0.20 m)2  4  4t  T/s  0.16 1  t  V  I   1.6 1  t  A
dt
dt
R
The magnetic field is increasing over the interval 0 s < t < 1 s and is decreasing over the interval 1 s < t < 2 s, so the
induced emf and current must have opposite signs in the second half of the time interval. We arbitrarily choose the
sign to be positive during the first half.
Time (s)
0.0
0.5
1.0
1.5
2.0
B (T)
0.00
1.50
2.00
1.50
0.00
E (volts)
0.16
0.08
0.00
–0.08
–0.16
I (A)
1.6
0.8
0.0
–0.8
–1.6
(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 s and t  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.
Assess: Notice in the graph how I  0 A at t  1 s, the instant in time when B is a maximum, that is, when dB/dt  0.
At this point the flux is (instantaneously) not changing so the corresponding induced emf and current are zero.
34.30. 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: (a)
(b) Since the field is perpendicular to the plane of the coil, A is parallel to B and   AB . The emf and current are
E(t ) 
d
dB
 NA
 100 (0.020 m) 2 1  12 t  T/s   0.126  (1  12 t ) V
dt
dt
I (t ) 
E(t )  0.126  (1  12 t ) V

  0.126  (1  12 t ) A
R
1.0 
(c) Using the expression for I(t) from part (b),
I 1 s    0.126 1  12 1 A  6.3 102 A  63 mA
Also, I  2 s  0.0 A and I 3 s   63 mA .
34.39. Model: Since the solenoid is fairly long compared to its diameter and the loop is located near the center,
assume the solenoid field is uniform inside and zero outside.
Visualize: Please refer to Figure P34.39. 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: (a) Using Faraday’s law, the induced current is
I loop 
Eloop
R

1 d  Asol dBsol  rsol2 d  0 NI 




R dt
R dt
R dt  l 

 rsol2 0 N dI
Rl
dt
We have used the fact that the field is approximately zero outside the solenoid, so the flux is confined to the area
Asol   rsol2 of the solenoid, not the larger area of the loop itself. The current is constant from 0 s  t  1 s so dI/dt  0 A/s
and Iloop  0 A during this interval, and I(0.5s) = 0.
(b) During the interval 1 s  t  2 s, the current is changing at the rate |dI/dt |  40 A/s, so the current during this
interval is
  0.010 m   4 107 T m/A  100 
2
I loop 
 0.10   0.10 m 
 40 A/s  =1.58 104A=158 A
Thus I(1.5 s) = 158 A.
(c) The current is constant from 2 s  t  3 s, so again dI/dt  0 A/s and Iloop  0 A so I(2.5s) = 0.
Assess: The induced current is proportional to the negative derivative of the solenoid current, and is zero when there
is a constant current in the solenoid.
34.40. Model: Since the solenoid is fairly long compared to its diameter and the coil is located near the center,
assume the solenoid field is uniform inside and zero outside.
Visualize: Please refer to Figure P34.40. 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. We have
I coil 
2
2
Ecoil 1
d  coil Ncoil rcoil
d  0 NI 
N coil rcoil
0 N dI
 N coil


 
R
R
dt
R
dt  l 
Rl
dt
2
We used the fact that the coil flux is confined to the area Acoil   rcoil
of the coil, not the larger area of the solenoid.
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
5  0.0050 m   4  107 T m/A  120 
2
I coil 
 0.10   0.080 m 
 50 A/s   3.7 104 A  0.37 mA
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 current will 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. That is, it is clockwise when seen from the left.
34.50. Model: Assume there is no resistance in the rails. If there is any resistance, it is accounted for by the
resistor.
Visualize: Please refer to Figure P34.50.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
E
B 2l 2v  0.50 T   0.10 m   0.50 m/s 
 Blv 
lB  

 6.25  104 N  6.3  10 4 N
 lB 
R
R
2.0 
 R 
2
Fpush  Fmag  IlB 
2
(b) The power is
P  Fv   6.25  104 N   0.50 m/s   3.1  104 W
(c) The flux is out of the page and decreasing and the induced current/field will oppose the change. The induced field
must have a flux that is out of the page so the current will be counterclockwise.
The magnitude of the current is
 Blv   0.50 T  0.10 m  0.50 m/s 
I 
 1.25 102 A

2.0 
 R 
(d) The power is
P  I 2 R  1.25 102 A   2.0    3.1104 W
2
Assess: From energy conservation we see that the mechanical energy put in by the pushing force shows up as
electrical energy in the resistor.