Download Physics I - Rose

Physics III
Homework III
CJ
Chapter 27; 76
Chapter 28; 82
Chapter 29; 7, 14, 20, 30, 50, 56, 61
27.76. IDENTIFY: The torque exerted by the magnetic field is     B. The torque required to
hold the loop in place is  .
SET UP:   IA.  is normal to the plane of the loop, with a direction given by the righthand rule that is illustrated in Figure 27.32 in the textbook.   IAB sin  , where  is the
angle between the normal to the loop and the direction of B.
EXECUTE:
(a) τ  IAB sin 60  (15.0 A)(0.060 m)(0.080 m)(0.48 T)sin60  0.030 N  m , in the  ˆj direction. To
keep the loop in place, you must provide a torque in the  ˆj direction.
(b) τ  IAB sin 30  (15.0 A)(0.60 m)(0.080 m)(0.48 T)sin30  0.017 N  m, in the  ˆj direction. You
must provide a torque in the  ˆj direction to keep the loop in place.
EVALUATE:
(c) If the loop was pivoted through its center, then there would be a torque on both sides
of the loop parallel to the rotation axis. However, the lever arm is only half as large, so
the total torque in each case is identical to the values found in parts (a) and (b).
28.82. IDENTIFY: Find the vector sum of the fields due to each sheet.
SET UP: Problem 28.81 shows that for an infinite sheet B  12 0 In . If I is out of the page,
B is to the left above the sheet and to the right below the sheet. If I is into the page, B is
to the right above the sheet and to the left below the sheet. B is independent of the
distance from the sheet. The directions of the two fields at points P, R and S are shown in
Figure 28.82.
EXECUTE:
(a) Above the two sheets, the fields cancel (since there is no dependence upon the
distance from the sheets).
(b) In between the sheets the two fields add up to yield
B  0nI , to
the right.
(c) Below the two sheets, their fields again cancel (since there is no dependence upon the
distance from the sheets).
EVALUATE: The two sheets with currents in opposite directions produce a uniform field
between the sheets and zero field outside the two sheets. This is analogous to the electric
field produced by large parallel sheets of charge of opposite sign.
Figure 28.82
29.7. IDENTIFY: Calculate the flux through the loop and apply Faraday’s law.
SET UP: To find the total flux integrate
d B
over the width of the loop. The magnetic
field of a long straight wire, at distance r from the wire, is B 
0 I
2 r
. The direction of
B
is
given by the right-hand rule.
EXECUTE:
(a) When B 
0i
2 r
(b) d  B  BdA 
, into the page.
0i
Ldr.
2 r
b
(c) B   a d B 
0iL b dr 0iL

ln(b/a).
2  a r
2
(d) E  d  B  0 L ln( b a ) di .
dt
(e) E 
2
dt
0 (0.240 m)
ln(0.360/0.120)(9.60 A/s)  5.06 107 V.
2
EVALUATE: The induced emf is proportional to the rate at which the current in the long
straight wire is changing
29.14. IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil.
SET UP: The flux through a coil is  = NBA cos  and the induced emf is
E  d /dt.
EXECUTE: The flux is constant in each case, so the induced emf is zero in all cases.
EVALUATE: Even though the coil is moving within the magnetic field and has flux
through it, this flux is not changing, so no emf is induced in the coil.
29.20. IDENTIFY: Use the results of Example 29.6. Use the three approaches specified in the
problem for determining the direction of the induced current. I  E/R .
SET UP:
Let
A be
directed into the figure, so a clockwise emf is positive.
EXECUTE:
(a) E  vBl  (5.0 m/s)(0.750 T)(1.50 m)  5.6 V
(b)
(i) Let q be a positive charge in the moving bar, as shown in Figure 29.20a. The magnetic
force on this charge is F = qv  B , which points upward. This force pushes the current in a
counterclockwise direction through the circuit.
is positive and is increasing in magnitude, so d  B / dt  0. Then by Faraday’s law
E  0 and the emf and induced current are counterclockwise.
(ii)
B
(iii) The flux through the circuit is increasing, so the induced current must cause a
magnetic field out of the paper to oppose this increase. Hence this current must flow in a
counterclockwise sense, as shown in Figure 29.20b.
(c)
E  RI . I 
E 5.6 V

 0.22 A.
R 25 
EVALUATE: All three methods agree on the direction of the induced current.
Figure 29.20
29.30. IDENTIFY: Use Eq.(29.10) to calculate the induced electric field E and use this E in
Eq.(29.9) to calculate E between two points.
(a) SET UP: Because of the axial symmetry and the absence of any electric charge, the
field lines are concentric circles.
(b) See Figure 29.30.
is tangent to the ring.
The direction of E (clockwise
or counterclockwise) is the
direction in which current will
be induced in the ring.
E
Figure 29.30
EXECUTE: Use the sign convention for Faraday’s law to deduce this direction. Let
be into the paper. Then
E 
dB
,E
dt
B
is positive. B decreasing then means
is positive and therefore clockwise. Thus
E
dB
dt
A
is negative, so by
is clockwise around the ring. To
calculate E apply úE  dl =  d B to a circular path that coincides with the ring.
dt
úE  dl = E(2 r)
 B  B r 2 ;
E (2 r )   r 2
d B
dB
  r2
dt
dt
dB
dB 1
and E  12 r
 2 (0.100 m)(0.0350 T/s)  1.75 103 V/m
dt
dt
(c) The induced emf has magnitude
E  úE  dl = E(2 r )  (1.75 103 V/m)(2 )(0.100 m)  1.100 103 V.
I
Then
3
E 1.100  10 V

 2.75  104 A.
R
4.00 
(d) Points a and b are separated by a distance around the ring of  r so
E  E ( r )  (1.75  103 V/m)( )(0.100 m)  5.50  104 V
(e) The ends are separated by a distance around the ring of 2 r so E  1.10 103 V as
calculated in part (c).
EVALUATE: The induced emf, calculated from Faraday’s law and used to calculate the
induced current, is associated with the induced electric field integrated around the total
circumference of the ring.
29.50. IDENTIFY: Apply Faraday’s law.
SET UP: For rotation about the y-axis the situation is the same as in Examples 29.4 and
29.5 and we can apply the results from those examples.
EXECUTE:
(a) Rotating about the y-axis: the flux is given by
Emax 
and
dB
  BA  (35.0 rad/s)(0.450 T)(6.00 102 m)  0.945 V.
dt
(b) Rotating about the x-axis:
d B
 0 and E  0.
dt
(c) Rotating about the z-axis: the flux is given by
Emax 
 B  BA cos
 B  BA cos
and
dB
  BA  (35.0 rad/s)(0.450 T)(6.00 102 m)  0.945 V.
dt
EVALUATE: The maximum emf is the same if the loop is rotated about an edge parallel
to the z-axis as it is when it is rotated about the z-axis.
29.56. IDENTIFY: Apply Newton’s 2nd law to the bar. The bar will experience a magnetic force
due to the induced current in the loop. Use a  dv / dt to solve for v. At the terminal speed,
a 0.
SET UP: The induced emf in the loop has a magnitude BLv . The induced emf is
counterclockwise, so it opposes the voltage of the battery, E.
EXECUTE:
(a) The net current in the loop is
ILB sin(90°) (E  BLv) LB
a F 

.
m
m
mR
I  E  BLv .
R
The acceleration of the bar is
To find v(t ) , set
dv  a  (E  BLv) LB
dt
mR
and solve for
v
using
the method of separation of variables:

v
0
t LB
2 2
dv
E

dt  v 
(1  e B L t /mR)  (10 m/s)(1  e t /3.1 s )
0
(E  BLv)
mR
BL
The graph of v versus t is sketched in Figure 29.56. Note that the graph of this function is
similar in appearance to that of a charging capacitor.
(b) Just after the switch is closed,
v  0 and I  E/R  2.4 A, F  ILB  2.88 N
and
a  F/m  3.2 m/s .
2
(c) When
v  2.0 m/s, a 
[12 V  (1.5 T)(0.8 m)(2.0 m/s)](0.8 m)(1.5 T)
 2.6 m/s  .
(0.90 kg)(5.0 )
(d) Note that as the speed increases, the acceleration decreases. The speed will
12 V
asymptotically approach the terminal speed E 
 10 m/s, which makes the
BL
(1.5 T)(0.8 m)
acceleration zero.
EVALUATE: The current in the circuit is counterclockwise and the magnetic force on
the bar is to the right. The energy that appears as kinetic energy of the moving bar is
supplied by the battery.
Figure 29.56
29.61. (a) and (b) IDENTIFY and Set Up:
The magnetic field of the wire is given by
I
B  0 and varies along the length
2 r
of the bar. At every point along the
bar B has direction into the page.
Divide the bar up into thin slices, as
shown in Figure 29.61a.
Figure 29.61a
EXECUTE: The emf dE induced in each slice is given by dE  v  B  dl . v  B is directed
I
d E  vB dr  v  0  dr. The total
 2 r 
b
d  L   Iv 
d  L dr

Iv
 Iv
d L
Vba   d E    0  dr   0 
  0 ln(r )d
a
d
d
2

r
2

r
2



0 Iv
 Iv
Vba  
(ln(d  L)  ln(d ))   0 ln(1  L/d )
2
2
toward the wire, so
emf induced in the bar is
EVALUATE: The minus sign means that Vba is negative, point a is at higher potential
than point b. (The force F = qv  B on positive charge carriers in the bar is towards a, so a
is at higher potential.) The potential difference increases when I or v increase, or d
decreases.
(c) IDENTIFY: Use Faraday’s law to calculate the induced emf.
SET UP: The wire and loop are sketched in Figure 29.61b.
EXECUTE: As the loop moves
to the right the magnetic
flux through it doesn’t
change. Thus E   d  B  0
dt
and I = 0.
Figure 29.61b
EVALUATE: This result can also be understood as follows. The induced emf in section
ab puts point a at higher potential; the induced emf in section dc puts point d at higher
potential. If you travel around the loop then these two induced emf’s sum to zero. There
is no emf in the loop and hence no current.