Download 07-HW1 - Rose

Physics III
Homework I
CJ
Chapter 32; 12, 24, 32, 35, 38, 46, 66, 68, 74
32.12. Model: The magnetic field is that of a current loop.
Solve:
From Example 32.5, the on-axis magnetic field of a current loop is
Bloop 
0

IR2
2 z 2  R2

3/ 2
 Bloop  z  R  
0

IR2
2 2 R2

3/ 2

0
I
2  2 3/ 2 R
Also, Bloop  z  0 m  0 I 2R. The ratio of these two fields is 23/ 2  0.354 .
32.24. Model: Assume that the solenoid is an ideal solenoid.
Solve:
We can use Equation 32.16 to find the current that will generate a 3.0 mT field inside the solenoid:
Bsolenoid 
0 NI
L
 I
Bsolenoid L
0 N
Using L  0.15 m and N  0.15 m 0.001 m  150 ,
I
 3.0  10
 4  10
3
7

T  0.15 m 

T m/A 150 
 2.39 A
Assess: This is a reasonable current to pass through a good conducting wire of diameter 1 mm.
32.32. Solve: The cyclotron frequency of a charged particle in a magnetic field is fcyc  qB/2m. An ion has charge q 
e. Its mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example, the mass of
N 2 is
m  mN  mN – melec 2(14.0031 u)(1.661  1027 kg/u) – 9.11  1031 kg  4.65174  1026 kg
Note that we’re given the atomic masses very accurately, and we need to retain that accuracy to tell the difference between
N 2 and CO. Calculating each mass and frequency:
Ion
Mass (kg)
N 2
4.65174  1026
1.6423

2
5.31341  10
4.64986  1026
1.4378
1.6429
O
f (MHz)
26

CO
Assess: The difference between N 2 and CO is not large but is easily detectable.
32.35. Model: Assume the magnetic field is uniform over the Hall probe.
Visualize: Please refer to Figure 32.41(a). The thickness is t  4.0  103 m.
Solve:
The Hall voltage is given by Equation 32.24:
VH 
15 A 1.0 T 
IB
IB
n

 2.9  1028 m 3
3
te

V
tne
1.0  10 m 1.60  10 19 C 3.2  10 6 V
H




Assess: The conduction electron density in metals is of the order of  5  1028 m3 (Table 28.1). The value
obtained for the charge carrier density is reasonable.
32.38. Model: Two parallel wires carrying currents in opposite directions exert repulsive magnetic forces on each
other. Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other.
Visualize: Please refer to Figure Ex32.38.
Solve:
The magnitudes of the various forces between the parallel wires are
F2 on 1 
7
0 LI1I2  2  10 T m/A   0.50 m 10 A 10 A 

 5.0  10 4 N  F2 on 3  F3 on 2  F1 on 2
2 d
0.02 m
F3 on 1
7
 0 LI1I 3  2  10 T m/A   0.50 m 10 A 10 A 


 2.5  10 4 N  F1 on 3
2 d
0.04 m
Now we can find the net force each wire exerts on the other as follows:





Fon 1  F2 on 1  F3 on 1  5.0  104 ˆj N  2.5  104 ˆj N  2.5  104 ˆj N  2.5  104 N, up





Fon 2  F1 on 2  F3 on 2  5.0  104 ˆj N  5.0  104 ˆj N  0 N





Fon 3  F1 on 3  F2 on 3  2.5  104 ˆj N  5.0  104 ˆj N  2.5  104 ˆj N  2.5  104 N, down

32.46. Model: The magnetic field is the superposition of the magnetic fields of three wire segments.
Visualize: Please refer to Figure P32.46
Solve: The magnetic field of the horizontal wire, with current I, encircles the wire. Because the dot is on the
axis of the wire, the input current creates no magnetic field at this point. The current divides at the junction, with I/2
traveling upward and I/2 traveling downward. The right-hand rule tells us that the upward current creates a field at the dot
that is into the page; the downward current creates a field that is out of the page. Although we could calculate the strength
of each field, the symmetry of the situation (the dot is the same distance and direction from the base of each wire) tells us
that the fields of the upward and downward current must have the same strength. Since they are in opposite directions,
their sum is 0 . Altogether, then, the field at the dot is B  0 T .
32.66. Model: A magnetic field exerts a magnetic force on a moving charge given by F  qv  B . Assume the
magnetic field is uniform.
Visualize: Please refer to Figure P32.66. The magnetic field points in the z-direction. If the charged particle is
moving along B , F  0 N . If v is perpendicular to B , the motion of the charged particle is a circle. However, when v
makes an angle with B , the motion of the charged particle is like a helix or a spiral. The perpendicular component of the
velocity is responsible for the circular motion, and the parallel component is responsible for the linear motion along the
magnetic field direction.
Solve:
From the figure we see that vy  v cos30 and vz  v sin30 . For the circular motion, the magnetic force
causes a centripetal acceleration. That is,
evy B 
mvy2
r
 r
mvy
eB
 9.11  10 kg  5.0  10 m/s cos30  0.82 mm
1.60  10 C  0.030 T 
31

6
19
The time for one revolution is
T


2 8.2  10 4 m
2 r

 1.2  10 9 s
vy
5.0  106 m/s cos30


The pitch p is the vertical distance covered in time T. We have




p  vzT  5.0  106 m/s sin30 1.2  10 9 s  3.0  10 3 m  3 mm
32.68. Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at
constant speed.
Solve:
The magnetic force on a proton causes the centripetal acceleration of
mv2
mv
 B
r
er
evB 
The maximum radius r is 0.25 m, and the desired speed v is 0.1c. So, the required field B is
1.67  10 kg  0.1  3  10 m/s  1.25 T
1.6  10 C  0.25 m 
27
B
8
19
32.74. Model: Assume that the 20 cm is large compared with the effective radius of the magnetic dipole (bar magnet).
Solve:
From Equation 32.9, the on-axis field of a magnetic dipole at a distance z from the dipole is
Bdipole 
 0 2
4 z3
The magnetic dipole moment can be obtained from the magnitude of the torque. We have
  B 

B sin

0.075 Nm
 0.5 T  sin90
 0.15 Am2
Thus, the magnetic field produced by the magnet is

Bdipole  10 7 T m/A


2 0.15 Am2
 0.20 m 
3
  3.75  10
6
T