Download Solutions #7

PHYSICS 172
WQ 2010
Solutions to Homework
#7
1. Giancoli Chapter 27, Problem 6
The magnetic force must be equal in magnitude to the force of gravity on the wire. The
maximum magnetic force is applicable since the wire is perpendicular to the magnetic field.
The mass of the wire is the density of copper times the volume of the wire.
FB  mg  I l B    12 d  l g 
2
3
3
3
2
 d 2 g  8.9  10 kg m   1.00  10 m   9.80 m s 
I

 1400 A
4B
4  5.0  105 T 
2
This answer does not seem feasible. The current is very large, and the resistive heating in
the thin copper wire would probably melt it.
2. Giancoli Chapter 27, Problem 8
We find the force per unit length from Eq. 27-3. Note that while the length is not known, the
direction is given, and so l  l ˆi.
F  I l  B  I l ˆi  B 
B
FB
l
 I ˆi  B   3.0 A 
ˆi
ˆj
kˆ
1
0
0
0.20 T

 1m 
 0.75ˆj  1.08kˆ N m 

 100 cm 

0.36 T 0.25T


  7.5 ˆj  11 kˆ  103 N cm
3. Giancoli Chapter 27, Problem 20
The velocity of each charged particle can be found using energy conservation. The electrical
potential energy of the particle becomes kinetic energy as it is accelerated. Then, since the
particle is moving perpendicularly to the magnetic field, the magnetic force will be a
maximum. That force will cause the ion to move in a circular path, and the radius can be
determined in terms of the mass and charge of the particle.
Einitial  Efinal  qV  12 mv 2  v 
Fmax  qvB  m
v
2
r
 r
mv
qB
m

2qV
m
2qV
1
m

qB
B
2mV
q
rd
rp
r
rp


1
B
2mdV
qd
1
B
2mpV
1
2mV
B
q
1
2mpV
B
qp
md
mp


qd
qp
qp

2
1
m
mp
q
qp

4
2
 2  rd  2 rp
 2  r  2 rp
4. Giancoli Chapter 27, Problem 25
The total force on the proton is given by the Lorentz equation, Eq. 27-7.
ˆi
ˆj

kˆ

FB  q  E  v  B   e   3.0ˆi  4.2ˆj  103 V m  6.0  103 m s 3.0  103 m s 5.0  103 m

0.45T
0.38T
0

 



 1.60  10 C   4.9ˆi  6.45ˆj  0.93kˆ    10 N C
  7.84  10 ˆi  1.03  10 ˆj  1.49  10 kˆ  N C
  0.78ˆi  1.0ˆj  0.15kˆ    10 N


s


 1.60  1019 C  3.0ˆi  4.2ˆj  1.9ˆi  2.25ˆj  0.93kˆ   103 N C
19
16
3
15
16
15
5. Giancoli Chapter 27, Problem 34
(a)
Since the velocity is perpendicular to the magnetic field,
the particle will follow a circular trajectory in the x-y plane of
radius r. The radius is found using the centripetal
acceleration.
mv 2
mv
qvB 
 r
r
qB
From the figure we see that the distance  is the chord distance, which is twice the
distance r cos . Since the velocity is perpendicular to the radial vector, the initial
direction and the angle  are complementary angles. The angles  and  are also
complementary angles, so   30.
2mv0
mv
  2r cos  
cos30  3 0
qB0
qB0
(b) From the diagram, we see that the particle travels a circular path, that is 2 short of a
complete circle. Since the angles  and  are complementary angles, so   60. The trajectory
distance is equal to the circumference of the circular path times the fraction of the complete
circle. Dividing the distance by the particle speed gives t.
l 2 r  360  2  60  2 mv0  2  4 m
t  



 v0 qB0  3  3qB0
v0
v0 
360

6. Giancoli Chapter 27, Problem 36
With the plane of the loop parallel to the magnetic field, the torque will be a maximum. We
use Eq. 27-9.

0.185m N
  NIAB sin   B 

 3.32 T
2
NIAB sin  1 4.20 A    0.0650 m  sin 90
7. Giancoli Chapter 27, Problem 44
Use Eq. 27-13.
 260V m 
q
E
 2 
 1.5  105 C kg
m B r  0.46T 2  0.0080m 
8. Giancoli Chapter 28, Problem 8
At the location of the compass, the magnetic field caused by the wire will point to the west,
and the Earth’s magnetic field points due North. The compass needle will point in the
direction of the NET magnetic field.
7
 I  4  10 T m A   43A 
Bwire  0 
 4.78  105 T
B net
2 r
2  0.18 m 
  tan
1
BEarth
Bwire
 tan
1
4.5  105 T
4.78  105 T
 43 N of W
B Earth

B wire
9. Giancoli Chapter 28, Problem 18
The magnetic field at the loop due to the long wire is into the page, and can be calculated by
Eq. 28-1. The force on the segment of the loop closest to the wire is towards the wire, since
the currents are in the same direction. The force on the segment of the loop farthest from the
wire is away from the wire, since the currents are in the opposite direction.
Because the magnetic field varies with distance, it is more difficult to calculate the total
force on the left and right segments of the loop. Using the right hand rule, the force on each
small piece of the left segment of wire is to the left, and the force on each small piece of the
right segment of wire is to the right. If left and right small pieces are chosen that are
equidistant from the long wire, the net force on those two small pieces is zero. Thus the total
force on the left and right segments of wire is zero, and so only the parallel segments need to
be considered in the calculation. Use Eq. 28-2.
Fnet  Fnear  Ffar 

 1
0 I 1 I 2
 II

1 
l near  0 1 2 l far  0 I1 I 2 l 


2 d near
2 d far
2
 d near d far 
4  107 T m A
2


6
  5.1  10 N, towards wire
0.030
m
0.080
m


 3.5A 2  0.100 m  
1

1
10. Giancoli Chapter 28, Problem 24
We break the current loop into the three branches of the triangle and add the forces from
each of the three branches. The current in the parallel branch flows in the same direction as
the long straight wire, so the force is attractive with magnitude given by Eq. 28-2.
 II 
F1  0 a
2 d
By symmetry the magnetic force for the other two segments will be equal. These two wires
can be broken down into infinitesimal segments, each with horizontal length dx. The net
force is found by integrating Eq. 28-2 over the side of the triangle. We set x=0 at the left
end of the left leg. The distance of a line segment to the wire is then given by r  d  3x .
Since the current in these segments flows opposite the direction of the current in the long
wire, the force will be repulsive.
a/2
a/2
0 II 
 II 
 II  
3a 
F2  
dx  0 ln d  3x
 0 ln 1 

0
0
2d 
2 3
2 3 
2 d  3x
We calculate the net force by summing the forces from the three segments.

F  F1  2 F2 



0 II 
 II  
3a  0 II   a
3 
3a  
a  2 0 ln 1 

ln 1 

 

2 d
2d 
  2d 3 
2d  
2 3 
11. Giancoli Chapter 28, Problem 31
Because of the cylindrical symmetry, the magnetic fields will be circular. In each case, we
can determine the magnetic field using Ampere’s law with concentric loops. The current
densities in the wires are given by the total current divided by the cross-sectional area.
I
I0
J inner  0 2
J outer  
2
 R1
  R3  R22 
(a) Inside the inner wire the enclosed current is determined by the
current density of the inner wire.
2
 B  ds  0 I encl  0  J inner R 
B  2 R   0
I 0 R 2
IR
 B 0 02
2
 R1
2 R1
(b) Between the wires the current enclosed is the current on the inner wire.
I
 B  ds  0 I encl  B  2 R   0 I 0  B  20 R0
(c) Inside the outer wire the current enclosed is the current from the inner wire and a
portion of the current from the outer wire.
 B  ds   I
0 encl
 0  I 0  J outer  R 2  R22  
2
2

  R 2  R22  
0 I 0  R3  R 
B  2 r   0  I 0  I 0

B


2 R  R32  R22 
  R32  R22  

(d)
(e)
Outside the outer wire the net current enclosed is zero.
 B  ds  0 I encl  0  B  2 R   0  B  0
3.0
See the adjacent graph.
2.5
1.5
-5
B (10 T)
2.0
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
2.5
R (cm)
12. Giancoli Chapter 28, Problem 37
(a) The magnetic field at point C can be obtained using the BiotSavart law (Eq. 28-5, integrated over the current). First break the loop into four
sections: 1) the upper semi-circle, 2) the lower semi-circle, 3) the right straight
segment, and 4) the left straight segment. The two straight segments do not contribute
to the magnetic field as the point C is in the same direction that the current is flowing.
Therefore, along these segments r̂ and d ˆ are parallel and d ˆ  rˆ  0 . For the upper
segment, each infinitesimal line segment is perpendicular to the constant magnitude
radial vector, so the magnetic field points downward with constant magnitude.
 I d ˆ  rˆ 0 I kˆ
I
B upper   0

 R1    0 kˆ .
2
2 
4 r
4 R1
4 R1
Along the lower segment, each infinitesimal line segment is also perpendicular to the
constant radial vector.
 I d ˆ  rˆ 0 I kˆ
I
Blower   0

 R2    0 kˆ
R1
4 r 2
4 R2 2
4 R2
C
Adding the two contributions yields the total magnetic field.
I
I
I
I 1 1 
B  B upper  Blower   0 kˆ  0 kˆ   0    kˆ
R2
4 R1
4 R2
4  R1 R2 
I
(b) The magnetic moment is the product of the area and the current. The
area is the sum of the two half circles. By the right-hand-rule, curling your fingers in the
direction of the current, the thumb points into the page, so the magnetic moment is in the k̂
direction.
  R2  R2 
 I 2
μ    1  2  Ikˆ 
R1  R22  kˆ

2 
2
 2
3.0
13. Giancoli Chapter 28, Problem 42
We treat the loop as consisting of 5 segments, The first has length
d, is located a distance d to the left of point P, and has current
flowing toward the right. The second has length d, is located a distance 2d to left of point P,
and has current flowing upward. The third has length d, is located a distance d to the left of
point P, and has current flowing downward. The fourth has length 2d, is located a distance d
below point P, and has current flowing toward the left. Note that the fourth segment is twice
as long as the actual fourth current. We therefore add a fifth line segment of length d,
located a distance d below point P with current flowing to the right. This fifth current
segment cancels the added portion, but allows us to use the results of Problem 41 in solving
this problem. Note that the first line points radially toward point P, and therefore by Problem
41(a) does not contribute to the net magnetic field. We add the contributions from the other
four segments, with the contribution in the positive z-direction if the current in the segment
appears to flows counterclockwise around the point P.
B  B 2  B3  B 4  B5


0 I
I
I
I
d
d
2d
d
kˆ  0
kˆ  0
kˆ  0
kˆ
1/
2
1/
2
1/
2
4  2d   4d 2  d 2 
4 d  d 2  d 2 
4 d  d 2  4d 2 
4 d  d 2  d 2 1/ 2
0 I 
5ˆ
 2 
k
4 d 
2 