Physics 122, Fall 2012
6 December 2012
Today in Physics 122: Examples in review
By class vote:
 Problem 22-40: offcenter charged
cylinders
 Problem 28-39: B
along
l
axis
i off
spinning, charged
disk
 Problem 30-74: selfinductance of a
toroid.
6 December 2012
Double rainbow over the VLA.
If you can explain everything in
this picture, you will have
understood all of electricity and
magnetism.
Physics 122, Fall 2012
1
Problem 22-40
A very long nonconducting cylinder of radius R1 is uniformly
charged with charge density E. It is surrounded by a
cylindrical metal (conducting) tube of inner radius R2 and
outer radius R3, which has no net charge. If the axes of the
two cylinders are parallel, but displaced from each other by a
distance d, determine the resulting electric field in the region
R > R3, where the radial distance r is measured from the
metal cylinder’s axis. Assume d < (R2 – R1).
6 December 2012
Physics 122, Fall 2012
2
Problem 22-40 (continued)
The key features or this problem
are that
(a) it involves a superposition of
two infinite cylindrical
charge distributions, so the
electric field has no
component in the direction
of the axes, suggesting the
use of Gauss’s Law; and
(b) E = 0 inside the conducting
cylindrical shell, also
suggesting the use of Gauss’s
Law.
6 December 2012
(c) University of Rochester
Physics 122, Fall 2012
R
R1
r
d
E
R2
R3
Gaussian cylinders of
length h, perpendicular to
the page.
3
1
Physics 122, Fall 2012
6 December 2012
Problem 22-40 (continued)
 The charge per unit length of
the nonconducting cylinder
is
1  E R12 .
Call that on the other
surfaces 2 and 3 .
 Draw a Gaussian cylinder,
radius r and length h, coaxial
with the conducting shell
and lying within it. E = 0
everywhere on its surface,
except the circular ends,
which are  E (i.e. E dA  0).
6 December 2012
R
R1
d
r
E
R2
R3
Gaussian cylinders of
length h, perpendicular to
the page.
Physics 122, Fall 2012
4
Problem 22-40 (continued)
 So Gauss’s Law for this
surface is
 1  2  h
1
Qencl 
EdA  0 

0
R
R1
0
 2  1  E R12
 The charge/length 2
is distributed nonuniformly
on the inner surface of the
shell, in such a way that its E
cancels that of the charged
rod, within the bounds of the
shell.
6 December 2012
d
r
E
R2
R3
Gaussian cylinders of
length h, perpendicular to
the page.
Physics 122, Fall 2012
5
Problem 22-40 (continued)
 The shell has no net charge,
so
3  2  E R12 .
 This charge must be
distributed uniformly on the
outer
t surface.
f
• Otherwise its E would
not be zero inside the
shell, as it must be since E
from the other two
charge distributions add
up to zero within the
shell.
6 December 2012
(c) University of Rochester
Physics 122, Fall 2012
R
R1
r
d
E
R2
R3
Gaussian cylinders of
length h, perpendicular to
the page.
6
2
Physics 122, Fall 2012
6 December 2012
Problem 22-40 (continued)
 Thus the calculation for the
field outside the shell is
familiar:
R
 EdA  Qencl  0
E2 Rh  3 h  0  E R12 h  0
E
E R12
2 0
R1
d
r
E
R2
rˆ .
 Note that the answer does
not depend on where the
charge rod is, as long as it’s
inside the shell.
6 December 2012
R3
Gaussian cylinders of
length h, perpendicular to
the page.
Physics 122, Fall 2012
7
Problem 28-39
A nonconducting circular disk, of radius R, carries a
uniformly-distributed electric charge Q. The plate is set
spinning with angular velocity  about an axis perpendicular
to the plate through its center. Determine
(a) its magnetic dipole moment and
(b) the
th magnetic
ti field
fi ld on it
its axis
i a di
distance
t
z away from
f
the
th
center.
0 
(c) Is B 
for z  R ?
2 z3
I’ll do (b) first.
6 December 2012
Physics 122, Fall 2012
8
Problem 28-39 (continued)
(b) We first break the problem
down into parts we have seen
before.
 Consider an infinitesimal
annulus, dr wide at radius r.
 This
Thi annulus
l carries
i a charge
h
dB
z
dQ   dA   2 rdr ,
and the disk is spinning, so the dr
r
annulus is like a circular loop
of current:
Charge Q , radius R
dQ 

dI 
dQ   rdr
Charge density   Q  R 2
2

6 December 2012
(c) University of Rochester
Physics 122, Fall 2012
9
3
Physics 122, Fall 2012
6 December 2012
Problem 28-39 (continued)
 That is a problem we’ve seen
before: in class on 30 October
2012 we showed that the
magnetic field on the axis of a
circular loop of radius R that
carries a current I is

R2 I
B 0
zˆ
32
2
R2  z2

dB
z

dB 
0
2
2
r
r dI
2
z
2

32
zˆ 
6 December 2012
0
2
r
dr
 So switch I  dI , R  r :
3
r
r dr
2
 z2

32
zˆ
Physics 122, Fall 2012
10
Problem 28-39 (continued)
 And so we add the contributions of all the annuli into
which the disk can be
decomposed:
B  zˆ
0
2
R

0
r
r 3 dr
2
 z2
dB
z

32
 Substitute:
u  r 2  z2
r
dr
du  2rdr
As r  0  R , u  z2  R 2  z2 , so...
6 December 2012
Physics 122, Fall 2012
11
Problem 28-39 (continued)
B  zˆ
 zˆ
0
4
R

0
r
r 2 2 rdr
2
 z2
 zˆ
0
4
R 2  z2

 u  z  du
u
z2
R2  z2

z2
 R 2  z2  z 
 z
2
2
2 

R z


  z 
1
2
1   R z  2 
 zˆ 0
2
2 
1   R z

 zˆ
2
32
u1 2 
 z2


 1 2  2
 1 2
z
0  u1 2
4

32
0 
6 December 2012
(c) University of Rochester
Physics 122, Fall 2012




12
4
Physics 122, Fall 2012
6 December 2012
Problem 28-39 (continued)
(a) The magnetic dipole moment
of the infinitesimal annulus at
radius r is its current times its
enclosed area:
dB
z
d   zˆ AdI  zˆ r 3 dr
so the sum of the moments of
all the annuli is just the
integral of this:
R
  zˆ  r 3 dr  zˆ
0
6 December 2012
r
dr
R4
4
Physics 122, Fall 2012
13
Problem 28-39 (continued)
(c) For this we need a power-series expansion that moves the
problem past the scope of PHY 122:

n!
xm

m
!
n
m
!


m0
 1  x n  
 1  nx 
n  n  1 2
x
2
if x  1.
2
Here we have in mind x   R z   1 and n   1 2 :
B
0 z 
2

 1   R z

6 December 2012

2 12

 2  1   R z
Physics 122, Fall 2012

2 1 2 


14
Problem 28-39 (continued)




2 12
2 1 2 
 2  1   R z
 1   R z



2
4
0 z  4    1  R  2 1  R  4
1R
3R 
 zˆ

 1        2  1       
4
2 R 
2 z 
8 z 
2 z 
8 z  


B  zˆ

0 z 
2
0 
2 z3
zˆ
So, yeah, it does what is suggested in the book.
6 December 2012
(c) University of Rochester
Physics 122, Fall 2012
15
5
Physics 122, Fall 2012
6 December 2012
Problem 30-74
(a) Show that the self-inductance L of a toroid of radius r0
containing N loops each of diameter d is
L
0 N 2 d 2
8r0
if r0  d.
(b) Calculate L for d = 2.0 cm and r0 = 66 cm. Assume that the
field inside the toroid is uniform, and that there are 550
loops in it.
6 December 2012
Physics 122, Fall 2012
16
Problem 30-74 (continued)
 Let’s calculate the field
inside the toroid first. The
circular geometry will
force B to be constant in
magnitude on circles and
point clockwise, so we use
Ampère’s Law and a
circular loop:
r0
 Bd  0 Iencl
B2 r  0 NI
B
0 NI ˆ

2 r
 
6 December 2012
Physics 122, Fall 2012
17
Problem 30-74 (continued)
 If d  r0 then all points
within the toroid lie at
approximately the same
radius: the field is almost
uniform.
 Thus the self-inductance
self inductance
becomes
N  B NBA
L

I
I
r0
2
 N2  d 
 N 2 d2
   0
 0
2 r0  2 
8r0
as advertised.
6 December 2012
(c) University of Rochester
,
Physics 122, Fall 2012
18
6
Physics 122, Fall 2012
6 December 2012
Problem 30-74 (continued)
 The arithmetic:
L  2.9  10 5 H.
r0
6 December 2012
(c) University of Rochester
Physics 122, Fall 2012
19
7