1 Solutions to Problem Set 9, Physics 370, Spring 2014

Solutions to Problem Set 9, Physics 370, Spring 2014
1
TOTAL POINTS POSSIBLE: 60 points.
~ at point P for
1. Griffiths Problem 5.09: Find the magnetic field B
each of the steady current configurations shown in Fig. 5.23 (below).
Figure 5.23
~ is a vector field, direction must be stated in the corHINT: Since B
rect answer. Also, for part (b), start by using the same general solution
for the magnetic field due to a segment of straight uniform current in
Example 5.5 you used to solve the Griffiths 5.08.
~ due to two different current configuWe are to find the magnetic field B
rations. We will tackle both by using the Biot-Savart law the magnetic
field due to a small segment of current I and length d~ℓ is
~ =
dB
µ0 d~ℓ × ˆ
.
I
2
4π
(1)
(a) (5 points possible) For the curved segment of radius a all points
on the wire are the same distance from P , so the field made by
the segment is just
Ba =
µ0 I
µ0 I (π/2)a
=
.
2
4π a
8a
(2)
This field points out of the page according to the right hand rule
(NOTE: Recall points from the current element to the loca~ field). The two straight segments
tion you are measuring the B
produce zero field at the point P because d~ℓ and point in the
same direction. The segment with radius b produces a field whose
2
Solutions to Problem Set 9, Physics 370, Spring 2014
magnitude is µ0 I/(8b) but is pointed into the page, so the total
field is
µ0 I 1 1
(out of the page);
(3)
−
B=
8
a b
because a < b the net field is positive, which is out of the page.
(b) (5 points possible) The field due to a straight wire segments is,
from Eq. (5.37) in the book,
B=
µ0 I
(sin θ2 − sin θ1 ).
4πs
(4)
For the bottom wire, θ1 = 0 and θ2 = π/2 so the field is
B = µ0 I/(4πR);
(5)
where the right hand rule gives a direction into the page for the
magnetic field. The field for the top straight wire is the same.
The curved segment has field
Bcurved =
µ0 I πR
,
4π R2
also into the page, so that the total field is
µ0 I 2
+ 1 (into the page)
B=
4R π
(6)
(7)
2. Griffiths Problem 5.16: Two long coaxial solenoids each carry current I, but in opposite directions, as shown in Fig. 5.42 (below). The
inner solenoid (radius a) has n1 turns per unit length, and the outer
~ in each of the three regions: (i) inside
one (radius b) has n2 . Find B
the inner solenoid, (ii) between them, and (iii) outside both.
Figure 5.42
3
Solutions to Problem Set 9, Physics 370, Spring 2014
(10 points possible) We start by considering the possible directions
for a magnetic field in a situation with cylindrical symmetry.
~ because, as argued in Ex• There can be no radial component to B,
ample 5.9, if there were a radial component, reversing the direction
of the current would switch this component. However, reversing
the current is the same as rotating the solenoid 180 degrees, which
should not alter a radial field.
~ be• There can be no significant circumferential component to B,
cause, again as argued in Example 5.9, any Amperian loop involving a circumferential line integral around the solenoids would in
fact enclose no significant currents.
~ field must be parallel to the axis of the solenoid. This
Therefore, any B
helps us select Amperian loops which will be easiest to work with by
~ The diagram
only using segments parallel to or perpendicular to B.
below shows the solenoids and two Amperian loops we will use to find
the field.
I
n2 turns/length
ℓ
ℓ
Loop 1
Loop 2
b
I
a
x
n1 turns/length
I
I
We will integrate counterclockwise around each loop. I’ll start with the
outside field first:
(iii) I didn’t sketch the loop used, but if you use a loop completely
outside the two solenoids, you find (as you did in example 5.9) that
the magnetic field must be the same everywhere outside the solenoid.
Since it must drop off to zero extremely far from the solenoid, it must
be zero everywhere outside both solenoids.
Solutions to Problem Set 9, Physics 370, Spring 2014
4
(i) To determine the magnetic field within the inner solenoid, I use
“Loop 1” with Amp´ere’s law
I
~ · d~ℓ = µ0 Ienc .
B
(8)
Taking the line integral for the loop, we can see that since the “top”
segment resides in a region with no magnetic field and the “right” and
”left” segments are perpendicular to any axial magnetic field, the only
segment of the wire which contributes to the integral is the “bottom
segment” such that
I
~ · d~ℓ = Bℓ.
B
(9)
Now for the right hand side. The current enclosed in Loop 1 is Ienc =
n2 ℓI − n1 ℓI = (n2 − n1 )ℓI.
Bℓ = (n2 − n1 )ℓI ⇒ B = µ0 (n2 − n1 )I
(10)
for the region s < a. With the current in the outer solenoid coming
out of the page, its contribution to the magnetic field will be in the
+ˆ
x whereas the inner solenoid will contribute a component in the −ˆ
x
direction, therefore:
~ = µ0 (n2 − n1 )I xˆ.
B
(11)
(ii) To determine the magnetic field between the inner and outer solenoid,
I use the Amperian loop labeled “Loop 2.” For Loop 2 the current enclosed is Ienc = n2 I and the integral of B is that same as before, so
B = µ0 n2 I, in the +ˆ
x direction.
By the way, you can do this whole problem without Amp´ere’s law if
you use the fact that the solenoid has magnetic field µ0 nI and just add
the magnetic fields of the solenoids like vectors.
(Thanks to Dr. Craig for providing the initial draft of this solution in
LATEX form.)
Solutions to Problem Set 9, Physics 370, Spring 2014
5
3. Griffiths Problem 5.20 (tweaked):
(a) Find the density ρ of mobile charges in a piece of copper, assuming each atom contributes one free electron. Useful Physical
gm
Constants: The atomic mass of copper, M = 64 mol
, the dengm
sity of copper is d = 9.0 cm3 , and Avagadro’s Number is A =
6.0 × 1023 /mol.
(b) Calculate the average electron velocity in copper wire 1mm in
diameter, carrying a current of 1 A.
(c) (1 pt Extra Credit) Your answer to part (b) should show the
average electron velocity is literally a snail’s pace. How, then, can
you carry on a long distance telephone conversation?
(d) Consider two such wires laying parallel to one another. Imagine
for a moment all the stationary positive ions (which are providing
charge neutrality) were removed. Compare the magnetic force
per unit length to the electrical force per unit length between two
such wires. In other words, write an expression showing how many
times greater than the electric force per unit length is compared to
the magnetic force per unit length, fe /fm . HINT: You worked out
the magnetic force per unit length between two current carrying
wires in Example 5.5 of the textbook. The electric field of a line
charge was worked out in Example 2.2 of the textbook. Work
out an expression for the electric force per unit length in terms
of currents and use the fact that the speed of light c = √ǫ10 µ0 to
allow you compare electric force per unit length to magnetic force
per unit length.
(a) (3 points possible) If we assume, as suggested, that each copper
atom contributes one free electron, then the number of copper
atoms per unit volume is the same as the number of free electrons
Solutions to Problem Set 9, Physics 370, Spring 2014
6
per unit volume. So we compute:
atoms mole gram
·
·
mole gram volume
1
=A·
·d
M
gm 1
23
9.0
= 6.02 × 10 /mol
gm
64 mol
cm3
ne =
= 8.47 × 1022 cm−3 .
(12a)
(12b)
(12c)
(12d)
Given the number density of electrons from equation 12d, the
charge density ρ is just this times the charge per electron, e =
1.6 × 10−19 C, or:
C
ρ = ene = 1.6 × 10−19 C 8.47 × 1022 cm−3 = 1.35 × 104 3 (13)
cm
(b) (3 points possible) The current per unit area perpendicular to
~ can be written (via equaiton 5.26):
flow, J,
J~ = ρ~v
(14)
We can relate this to the current (which we were given) because we
are told this is a copper wire. Assuming a circular cross-section,
which means for a wire of radius a:
J=
I
πa2
(15)
and so:
I
1A
I
−3 cm
=
ρv
→
v
=
=
.
=
9.1×10
C
πa2
ρπa2
s
1.35 × 104 cm3 π(0.05cm)2
(16)
Wow, this is only 0.0091 cm/s or about 33 cm/hour!
(c) (1 extra credit point possible) While the “drift velocity” of
the free charges in a wire is quite low, what is “carrying” the information about the conversation you are having on the phone is
an electromagnetic wave. This electromagnetic wave travels at a
speed which is much faster that the individual speed of the elec~ field change propagates much faster than the
trons. Since this E
Solutions to Problem Set 9, Physics 370, Spring 2014
7
~ far down the wire changes, causing the
electrons themselves, the E
electrons there to respond... thus a phone conversation traveling
much faster than the electrons in the wire.
(d) (4 points possible) From equation (5.40), the force per unit
length between these two wires is:
fm =
µ 0 I1 I2
.
2π d
(17)
Now, if we instead consider the electric force between two wires
where all the bound charges were removed, then starting with
equation (2.9) which gives the electric field near a wire:
E=
1 2λ
4πǫ0 s
(18)
which allows me to write the electric force per unit length of the
wire for two wires separated by distance d as:
fe =
qE
1 2λ1 λ2
F
=
= λE =
.
ℓ
ℓ
4πǫ0 d
(19)
I can relate the free line charge densities, λ to current because
I = λv (equation 5.13), therefore λ = I/v. I can also note that
c2 = ǫ01µ0 → ǫ0 = c21µ0 and thus I can rewrite equation 19 as
1 2I1 I2
c2 µ0 2I1 I2
c2 µ 0 I1 I2
c2
fe =
=
= 2
= 2 fm ,
4πǫ0 v 2 d
4π v 2 d
v 2π d
v
(20)
And thus:
fe
c2
(3 × 1010 cm/s)2
= 2 =
= 1.1 × 1025
fm
v
(0.0091cm/s)2
(21)
Wow, electrical forces are much stronger than magnetic forces unless
charge neutrality exists!
Solutions to Problem Set 9, Physics 370, Spring 2014
8
4. Griffiths Problem 5.47: The magnetic field on the axis of a circular
R2
current loop, B(z) = µ20 I (R2 +z
(equation 5.41). is far from uniform
2 )3/2
(it falls off sharply with increasing z). You can produce a more nearly
uniform field by using two such loops a distance d apart (See Figure
5.60, below).
Figure 5.60
(a) Find the field (B) as a function of z, and show that ∂B
is zero
∂z
at the point midway between them (z = 0). Now, if you pick
d just right the second derivative of B will also vanish at the
midpoint. This arrangement is known as the Helmholtz coil;
it’s a convenient way of producing relatively uniform fields in the
laboratory.
2
(b) Determine d such that ∂∂zB2 = 0 at the midpoint, and find the
√0 I ]
resulting magnetic field at the center. [Answer : 58µ
5R
(a) (6 points possible) The magnetic field at the center of two loops
as illustrated in Figure 5.60 should just consist of the vector sum of
the magnetic field a distance d2 below one such loop with another
idential loop a distance d2 above such a loop. Equation 5.41 gives
the magnetic field anywhere along the axis of a loop:
µ0 I
R2
~
B(z)
=
zˆ.
2 (R2 + z 2 )3/2
(22)
9
Solutions to Problem Set 9, Physics 370, Spring 2014
If we place the two loops at z = − d2 and z = d2 then the distance
for some point on the z-axis from the currents is d2 = z and d2 − z
respectively, therefore we can write the expression for the magnetic
field on the axis due to the two loops as:
~
~ due to lower ring + B
~ due to upper ring
B(z)
=B
(23a)
#
"
µ0 I
R2
R2
=
+
3/2 zˆ (23b)
3/2
2
R2 + ( d2 + z)2
R2 + ( d2 − z)2
#
"
µ0 IR2
1
1
=
3/2 +
3/2 zˆ.
2
R2 + ( d + z)2
R2 + ( d − z)2
2
2
(23c)
Now all I need to do is find the derivative and evaluate it at z = 0,
which is just an annoying application of the chain rule:
#
"
∂B
1
µ0 IR2 ∂
1
=
+
3/2
∂z
2 ∂z R2 + ( d + z)2 3/2
R2 + ( d − z)2
2
2
(24a)
−5/2
d
d
µ0 IR2 −3
2
2
R + ( + z)
+z
2
=
2
2
2
2

2 !−5/2 −3
d
d
+
R2 +
−z
− z (−1) (24b)
2
2
2
2


d
− d2 + z
−z
3µ0 IR2 

2
=
+

5/2 5/2 
2
2
2
R2 + d2 + z
R2 + d2 − z
"
(24c)
#
"
d
− d2
√
∂B 3µ0 IR2
2
=0
+
=
5/2
5/2
∂z z=0
2
R2 + ( d2 )2
R2 + ( d2 )2
(24d)
As indicated by equation 24d, the derivative at the origin is zero,
as expected.
10
Solutions to Problem Set 9, Physics 370, Spring 2014
(b) (4 points possible) All we need here is to evaluate the second
derivative and determine for what values of d it goes to zero at the
origin as well. Applying the chain rule to derivative of equation
24c, we find:
∂2B
∂ ∂B
=
(25a)
2
∂z
∂z ∂z
#
"
d
−
z
− d2 + z
3µ0 IR2 ∂
2
=
+
5/2
2
∂z R2 + ( d + z)2 5/2
2
R + ( d2 − z)2
2
(25b)

d
− d2 + z −5
2
+
z
−1
3µ0 IR2 
2
2
=

5/2 + 2
2 7/2
2
R2 + d2 + z
R2 + d2 + z

d
− z −5
2 d2 − z (−1) 
−1
2
2
+
+ 2 5/2
2 7/2 
d
d
2
2
R + 2 −z
R + 2 −z
(25c)

3µ0 IR2 
=

2
5
+ R2 +
d
2
d
2
−2
R2 +
−z
2
−z
d
2
+z


2 7/2 
5
2 5/2 + R2 +
d
2
d
2
+z
2
+z
2 7/2
(25d)
11
Solutions to Problem Set 9, Physics 370, Spring 2014
And now, evaluating this at the origin:

2
5 d2
3µ0 IR2 
−2
∂ 2 B =

5/2 + 7/2
∂z 2 z=0
2
d 2
d 2
2
2
R + 2
R + 2

2
5 d2

(26a)
+ 7/2 
2
R2 + d2
"
2 !
2 #
2
3µ0 IR
d
d
2
= + 10
7/2 −2 R +
2
2
2
2 R2 + d2
(26b)
3µ0 IR2
=
2 7/2
R2 + d2
" 2
d
−
5
2
2
3µ0 IR2
2
d
−
R
=
2 7/2
R2 + d2
2 !#
d
R2 +
2
(26c)
(26d)
This second derivative will go to zero when d2 − R2 → 0 or d = R,
in which case the magnetic field at the origin is (using equation
23c):
#
"
2
µ
IR
1
1
0
~
B(0)
=
(27a)
3/2 +
3/2 zˆ
2
R2 + ( R2 )2
R2 + ( R2 )2
"
#
1
= µ0 IR2
zˆ
(27b)
2 3/2
5R
4
µ0 IR2 43 2
53/2 R3
8µ0 I
= 3/2
5 R
=
as noted in the wording of the problem.
(27c)
(27d)
Solutions to Problem Set 9, Physics 370, Spring 2014
12
5. Griffiths Problem 5.21 (extensively modified): Recall the general
rule (expressed in equation 1.46) that the divergence of a curl is always
zero.
~ ×B
~ = µ0 J~ and
(a) Using the differential form of Amp´ere’s law, ∇
∂ρ
~
~
the continuity equation ∇ · J = − ∂t to write an expression for
~ · (∇
~ × B).
~
∇
(b) At first blush, the expression you computed in part (a) doesn’t
equal zero... Show that differential form of Amp´ere’s law you used
in part (a) can not be valid except in the realm of magnetostatics (HINT: Read section 5.2.1 for a discussion of the conditions
necessary for a steady current as necessary in magnetostatics).
(c) Verify that there is no other such problems taking the curl of any
divergences in the remaining Maxwell equations and the divergence of any curls in the remaining Maxwell equations (as listed
on page 241 of your textbook).
(a) (3 points possible) Taking the divergence of the differential form
of Amp´ere’s law we have:
~ · (∇
~ × B)
~ = µ0 ∇
~ · J~ = −µ0 ∂ρ
∇
∂t
(28)
which as I noted, at first blush, doesn’t look a lot like zero.
(b) (4 points possible) For equation 28 to equal zero requires ρ to
= 0), which is in fact one of the conditions
be a constant ( ∂ρ
∂t
for a steady current, which is a requirement for magnetostatics. I
will note that this suggests when we are in the realm of magnetodynamics (varying currents), then it is unlikely the curl of the
~ ×B
~ = µ0 J.
~
magnetic field will simply be Amp´ere’s law, ∇
(c) (3 points possible) Let’s check the divergence of the curl of the
~ field:
E
~ · (∇
~ × E)
~ =∇
~ · (0) = 0
∇
(29)
Solutions to Problem Set 9, Physics 370, Spring 2014
13
So that obeys the general rule that the divergence of a curl is
always zero. How about the curls of the other two Maxwell equations as listed on page 241 of the textbook?
~ × (∇
~ · B)
~ =∇
~ × (0) = (undefined)
∇
~ × (∇
~ · E)
~ =∇
~ × ρ = (undefined)
∇
ǫ0
(30a)
(30b)
but these are not a problem because there is no relevant second
derivative expression that requires curl(div) = 0.
6. Griffiths Problem 5.30: Use the results of Example 5.11 (p. 245
~ field inside a uniformly charged sphere, of total
–247) to find the B
charge Q and radius R, which is rotating at a constant angular velocity ω.
(10 points possible) The major result from Example 5.11 we want
to start with is the magnetic vector potential of a rotating uniformly
charged spherical shell given by equation (5.69):
(
µ0 Rωσ
ˆ r<R
r sin θφ,
3
~ r) =
(31)
A(~
4
µ0 R ωσ sin θ ˆ
φ, r ≥ R
3
r2
Notice I said this is the expression of the vector potential of a rotating
uniformly charged spherical shell. To get the equivalent vector for a
rotating uniformly charge sphere, we have to integrate across the contribution to the vector potential of all the shells that make up a sphere.
This is fairly simple as long as we properly distinguish the unprimed
coordinates (where we are evaluating A) and the primed coordinates
(where the current resides and where we have to integrate over).
If we take σ = ρdr ′ , then at a given distance from the center of the
sphere, we have to consider the contribution to the vector potential
from shells at radius r ′ inside r using the “outside” vector potential
expression
′4
~ r′ >r = µ0 ρr ω sin θ dr ′φˆ
(32)
dA
3
r2
where I substitute r ′ for R because R represented the radius of the shell
contributing to the vector potential. The contribution to the vector
14
Solutions to Problem Set 9, Physics 370, Spring 2014
potential from shells at radius r ′ outside a specific radius r using the
“inside” vector potential expression:
µ0 ρr ′ ω
~
′
r sin θdr ′φˆ
dAr <r =
3
(33)
where R is the radius of the sphere in this case. Thus the total vector
potential expression a distance r from the center is:
~=
A
Zr
~ r′ <r +
dA
r ′ =0

=
µ0 ρω sin θ
3
r2
ZR
r ′ =r
Zr
r ′ =0
5
~ r′ >r
dA
r ′4 dr ′ +
(34a)
µ0 ρω
r sin θ
3
2
µ0 ρω
R −r
µ0 ρω sin θ r
+
r sin θ
2
3
r 5
3
2
2
2
2
R −r ˆ
µ0 ρωr sin θ r
φ
+
=
3
5
2
2
2
R
3r
µ
ρωr
sin
θ
0
~=
φˆ
−
A
3
2
10
=
ZR
r ′ =r
2
φˆ

r ′ dr ′  φˆ
(34b)
(34c)
(34d)
(34e)
~ we simply need to solve for the
Now, to solve for the magnetic field B,
~ as expressed in equation 34e in
curl of the magnetic vector potential A
Solutions to Problem Set 9, Physics 370, Spring 2014
15
spherical coordinates (noting only Aφ is non-zero):
~ =∇
~ ×A
~
B
(35a)
1 ∂
1 ∂
(sin θAφ )ˆ
r−
(rAφ )θˆ
(35b)
=
r sin θ ∂θ
r
∂r
2
1 µ0 ρωr R
3r 2 ∂
µ0 ρω sin θ 1 ∂ r 2 R2 3r 4 ˆ
2
=
−
(sin θ)ˆ
r−
−
θ
r sin θ 3
2
10 ∂θ
3
r ∂r
2
10
(35c)
2
2
µ0 ρω sin θ 1 2rR2 12r 3 ˆ
3r
1 µ0 ρωr R
(2 cos θ sin θ)ˆ
r−
−
−
θ
=
r sin θ 3
2
10
3
r
2
10
(35d)
2
µ0 ρω R
µ0 ρω sin θ 2R2 12r 2 ˆ
3r 2
=
(2 cos θ)ˆ
r−
−
−
θ (35e)
3
2
10
3
2
10
2
2
R
R
3r 2
6r 2
2µ0 ρω
ˆ
~
cos θˆ
r−
sin θθ
(35f)
−
−
B=
3
2
10
2
10
3Q
,
4πR3
therefore equation 35f becomes:
2
2
2
2
µ
Qω
R
R
3r
6r
0
~ =
B
cos θˆ
r−
sin θθˆ
−
−
2πR3
2
10
2
10
But ρ =
(36a)