Download PHYS 110B - HW #3

PHYS 110B - HW #3
Spring 2004, Solutions by David Pace
Any referenced equations are from Griffiths
Problem statements are paraphrased
[1.] Problem 7.20 from Griffiths
Two circular loops of wire share the same axis but are displaced vertically by a distance,
z. Reference figure 7.36 in Griffiths. The wire of radius a is considerably smaller than the
wire of radius b.
(a) The larger loop (of radius b) carries a current I. What is the magnetic flux through
the smaller loop due to the larger? (Griffiths’ Hint: The field of the large loop may be
considered constant in the region of the smaller loop.)
(b) If the same current I now flows in the smaller loop, then what is the magnetic flux
through the larger loop? (Griffiths’ Hint: The field of the smaller loop may be treated as a
dipole.)
(c) What is the mutual inductance of this system? Show that M12 = M21 .
Solution
(a) Let the smaller loop be 1 and the larger loop 2. The flux through loop 1 is given by,
Z
~ 2 · d~a1
Φ1 = B
(1)
Griffiths’ hint is that the field of the larger loop is constant in the region of the smaller loop.
The magnetic field of a circular loop, a distance z along the axis, is given in example 5.6 of
the text as,
R2
µo I
Eq. 5.38
(2)
B(z) =
2 (R2 + z 2 )3/2
where R = b in this problem.
Since this field is constant in the region of the smaller loop, the flux is given by this field
multiplied by the area of the smaller loop.
Φ1
=
B2 · πa2
=
µo πa2 b2 I
2(b2 + z 2 )3/2
(3)
(b) In this problem we can greatly simplify the algebra by choosing a particular surface
through which to calculate the flux. Recall that the flux through the larger loop is,
Z
~ 1 · d~a2
Φ2 = B
(4)
The integral in (4) may be taken over any surface that is bounded by the loop through
which we want to know the flux. Notice that there is a form of the dipole magnetic field
that has specific spherical coordinate representation,
~ dip = µo m 2 cos θ r̂ + sin θ θ̂
(5)
B
4πr3
1
where m is the magnetic moment of the small loop. Recall from equation 5.84 that m =
Iπa2 .
Now if we can define a surface that is directed entirely along either the r̂ or θ̂ direction,
then the dot product in (4) will simplify quite well. This surface must also be bounded by
the larger loop since this is the loop through which we desire to know the flux.
Defining a spherical coordinate system with the smaller loop at the center it is seen that
a surface in the r̂ direction exists by sweeping the vector from the small loop through an
angle θ0 and a complete 2π revolution in the φ̂ direction.
Figure 1: Geometry of the suggested method for determining the flux in part (b). A spherical
coordinate system is centered on the smaller loop (approximated as a dipole, m)
~ and the surface
used in the flux calculation is that spherical section traced out by the vector ~r and bounded on
the larger loop.
The flux through 2 is,
Z
θ0
2π
Z
Φ2 =
0
0
Z
= 2π
0
θ0
µo m 2
cos
θ
r̂
+
sin
θ
θ̂
· r2 sin θ dθ dφ r̂
4πr3
µo Iπa2
(2 cos θ)r2 sin θ dθ
4πr3
(6)
(7)
The distance r needs to be specified in terms of given parameters. Geometry allows both r
and θ0 to be written in terms of z and b.
√
r =
b2 + z 2
(8)
sin θ0 =
b
r
2
=
√
b
b2 + z 2
(9)
Continue solving for the flux through loop 2,
θ0
µo Ia2
(cos θ) sin θ dθ
2r
0
Z θ0
µo πIa2
= √
(cos θ) sin θ dθ
b2 + z 2 0
θ 0
µo πIa2
1 2
= √
sin θ
b2 + z 2 2
0
µo πIa2
b2
= √
2 b2 + z 2 b2 + z 2
Z
Φ2 = 2π
µo πa2 b2 I
=
2(b2 + z 2 )3/2
(10)
(11)
(12)
(13)
(14)
(c) The mutual inductance, M , of two loops of wire is given by,
Φ1 = M12 I2
Φ2 = M21 I1
(15)
where the subscripts denote the wire. In this problem the currents through the wires are
the same, I1 = I2 .
Finding M12 ,
M12
=
Φ1
I
=
Φ2
I
=
µo πa2 b2
2(b2 + z 2 )3/2
(16)
=
µo πa2 b2
2(b2 + z 2 )3/2
(17)
For M21 ,
M21
From (16) and (17) we confirm that the mutual inductances are equivalent.
[2.] Problem 7.21 from Griffiths
Two long wires are coplanar and a distance 3a away from each other (these may be treated
as a large rectangular loop in which the other two sides are very far away and should be
neglected). In between these wires and also coplanar is a square loop of wire. This square
loop has sides of length a. Find the emf in the larger loop when a clockwise current, I,
flows in the smaller loop. This current has a time dependence given by dI/dt = k, where k
is a constant. What is the direction of the current induced in the larger loop?
Solution
The emf induced in any loop due to a magnetic field is,
E =−
dΦ
dt
(18)
where Φ is the magnetic flux through the loop in question.
It is difficult to calculate the magnetic flux through the larger loop due to the smaller because the magnetic field of the smaller loop is irregular. Finding the magnetic field only at
3
the center of a square loop is the goal of problem 5.8 in Griffiths, and any determination of
the field beyond this simplification is time consuming.
Consider two loops of wire. One of the most important points concerning inductance is
that the flux through loop one due to the current in loop two is the same as the flux through
loop two due to the same current in loop one. This problem is more easily solved by placing
the current I in the larger loop and finding the flux through the smaller.
Put a clockwise current in the larger loop. Let the left wire have current going up the page
and the right wire have current going down the page. The flux through the smaller loop is
found by using the generalized form of (1). Due to the symmetry in the problem it is noted
that each of the wires makes the same contribution to the total flux through the smaller
loop. The total flux may be calculated by finding the flux due to one of the wires and
multiplying this value by 2.
The magnetic field a distance s away from an infinite (or just really long) wire carrying
current I is,
~ = µo I φ̂
(19)
B
2πs
Here the field of each wire points into the page inside the smaller loop. Using cylindrical
coordinates the flux through the smaller loop is,
Z
2a
a
Z
B ds dz
Φ = 2
a
Z
2a
ds
s
a
2a
µo aI
ln
=
π
a
µo aI
=
π
=
(20)
0
µo aI
ln(2)
π
(21)
(22)
(23)
Therefore, in the case of current I in the smaller loop, the flux through the larger loop is
given by,
µo aI
ln(2)
(24)
Φ=
π
The emf induced in this loop is,
d µo aI
E = −
ln(2)
dt
π
(25)
= −
µo a
dI
ln(2)
π
dt
(26)
= −
µo a
k ln(2)
π
(27)
The flux calculation was done with reference to the direction into the page. Since the resultant emf value is negative the direction of the resultant current is found with reference
4
to the direction out of the page. Placing one’s right thumb in the out of the page direction
gives a current (along one’s fingers) in the counter-clockwise direction.
This direction could also have been determined using Lenz’s law early in the solution.
Since the current in the wires is increasing, the flux into the page through the smaller loop
is also increasing. The induced current in the smaller loop seeks to combat this increase
and must therefore be directed counter-clockwise such that its field is out of the page.
From properties of induction it is known that putting the same current through the smaller
loop results in the same induced behavior of the larger loop.
[3.] Problem 7.24 from Griffiths
A time dependent current, I = Io cos(ωt), flows in a straight wire. This wire runs along the
axis of a toroidal coil with rectangular cross section that is connected to a resistor, R. Use
the following values for various parameters:
Io = 0.5 A
ω = 2πf = 120π Hz
Toroid Inner Radius, a = 0.01 m
R = 500 Ω
Toroid Outer Radius, b = 0.02 m
Toroid Height, h = 0.01 m
Toroid No. of Turns, N = 1000
(a) What is the emf in the toroid? What is the current in the resistor due to this emf? Treat
this system as quasistatic.
(b) Find the back emf in the toroid. This is due to the current found in (a). Calculate the
ratio of this emf to the one from part (a).
Solution
(a) There is an equal amount of flux through each loop of the toroid. Since the magnetic
field due to the wire is known we can solve for the flux through one loop of the toroid
immediately,
Z bZ
h
µo I
ds dz
a
0 2πs
µo hIo cos(ωt)
b
ln
=
2π
a
Φ =
(28)
(29)
(4π × 10−7 )(0.01)(0.5) cos(ωt)
=
ln(2)
2π
(30)
= 6.93 × 10−10 cos(120πt)
(31)
where magnetic flux is in units of T · m2 .
The total flux through the toroid is this value multiplied by the number of turns.
Φtot
=
(1000)Φ
=
5
6.93 × 10−7 cos(120πt)
(32)
The emf in the toroid is found using (18),
d
6.93 × 10−7 cos(120πt)
dt
(33)
= −6.93 × 10−7 (−120π) sin(120πt)
(34)
= 2.61 × 10−4 sin(120πt)
(35)
E = −
where emf is in units of Volts.
The current in the resistor is,
I
=
E
R
5.22 × 10−7 sin(120πt)
=
(36)
in units of Amperes.
(b) The back emf, Eb , of a system is that due to its self-inductance.
Eb = −L
dI
dt
(37)
where L is the self-inductance of the system and I is the current in the system.
We have already solved for the current in the toroid, and the self-inductance is given in
example 7.11 from Griffiths.
µo N 2 h
b
L=
ln
Eq. 7.27
(38)
2π
a
The back emf is,
Eb
µo N 2 h
ln
= −
2π
= −
b
d
−7
5.22 × 10 sin(120πt)
a
dt
(4π × 10−7 )(1000)2 (0.01)
ln(2)(5.22 × 10−7 )(120π) cos(120πt)
2π
= −2.73 × 10−7 cos(120πt)
(39)
(40)
(41)
and the sign is inconsequential here.
The ratio of amplitudes of the back emf to the primary emf is (absolute value),
Eb
E
=
2.73 × 10−7
2.61 × 10−4
=
1.05 × 10−3
(42)
The back emf is approximately 1000 times smaller than the emf directly induced by the
center wire.
6
[4.] Problem 7.25 from Griffiths
Reference the circuit diagram in figure 7.38 of Griffiths. The capacitor, C, is charged to
potential V and then connected to the inductor L. The switch is closed at time t = 0.
Determine the current, I = I(t). What happens if a resistor, R, is connected in series to this
circuit?
Solution
Both the capacitor and inductor serve to establish a potential through the circuit (see footnote 11 in Griffiths on page 315 for further comment) so we have,
(43)
Vcap + Vind = 0
Q
dI
−L
C
dt
= 0
(44)
dI
Q
= L
C
dt
(45)
In this problem the capacitor is discharging so we may write the current as,
I=−
dQ
dt
(46)
This gives,
d
Q
= L
C
dt
−
d2 Q
1
Q =
LC
dt2
dQ
−
dt
(47)
(48)
The solution to this differential equation is known and used frequently in electrodynamics
(see equation 3.26 and the corresponding section about the Separation of Variables technique in Griffiths).
t
t
+ B cos √
(49)
Q(t) = A sin √
LC
LC
where A and B are constants to be determined from the boundary conditions.
The boundary conditions here are time constraints instead of the usual physical constraints,
but the method for finding the constants is unchanged. At time t = 0 the charge on the capacitor is known to be CV .
0
0
Q(t = 0) = CV = A sin √
+ B cos √
(50)
LC
LC
CV
= B
7
(51)
The initial current, before the switch is closed, is also zero.
I(t = 0)
dQ(t = 0)
dt
0
0
+ CV sin √
= −A cos √
LC
LC
0 = −
=
(52)
(53)
(54)
0 = A
The current is,
I(t)
=
d
t
dQ
= − CV cos √
−
dt
dt
LC
r
C
t
= V
sin √
L
LC
(55)
(56)
where the sign of the current does not matter because the sine term means it will oscillate.
This ideal circuit will oscillate forever.
The addition of a resistor will change the right side of (44) as follows,
dI
Q
− L = IR
C
dt
(57)
The resultant differential equation in this case is,
L
L
d2 Q Q
dQ
+
= −R
dt
C
dt
d2 Q
dQ Q
+R
+
= 0
dt
dt
C
(58)
(59)
This is also a well known differential equation and in the most general case we expect to get
solutions that are examples of damped oscillating systems. This circuit would eventually
stop oscillating because all of the energy stored between the capacitor and inductor would
be dissipated as heat in the resistor.
[5.] Problem 7.26 from Griffiths
Find the stored energy in a length l of a long solenoid using the methods listed below. The
solenoid has radius R, current I, and n turns per unit length.
(a) Use equation 7.29. The self-inductance is found in problem 7.22. (Carter Hint: Problem
7.22 was done in lecture on 19-4-2004.)
(b) Use equation 7.30. The vector potential is given in example 5.12.
(c) Use equation 7.34.
(d) Use equation 7.33. Use the cylindrical tube that extends from a < R to b > R as your
volume.
8
Solution
(a) We are supposed to use,
1
W = LI 2
2
Eq. 7.29
(60)
The current is given in the problem and from the lecture noted above we know the selfinductance of this solenoid is,
L = µo πlR2 n2
(61)
Therefore, the energy stored in this system is,
1
W = µo πlR2 n2 I 2
2
(62)
(b) The equation to use is,
1
W =
2
I
~ · I)
~ dl
(A
Eq. 7.30
(63)
From example 5.12 (page 238) the vector potential is,
~ = µo nI s φ̂
A
2
s<R
2
~ = µo nI R φ̂
A
2 s
s>R
(64)
(65)
The integrand in (63) is zero everywhere that the current is zero. The only place where this
is not zero is at the solenoid wires, i.e. for s = R. Notice that the expressions for vector
potential are equivalent at this location.
Another important issue is that (63) will return the energy due to only one of the solenoid
loops in its present form. The integrand is non-zero for each individual turn. Multiplying
this value by the total number of turns will give the total solenoid energy.
The integration is carried out over the length through which the current flows. This is in
the φ̂ direction and means that dl = s dφ. Using s = R and multiplying by the total number
of turns, N = nl, the energy of the solenoid is,
W
=
nl
2
I
~ · I)
~ dl = nl
(A
2
(c) Now use,
1
W =
2µo
Z
9
0
µo nI
R · I(Rdφ)
2
(66)
=
nl µo nI
RI(R2π)
2 2
(67)
=
1
µo πlR2 n2 I 2
2
(68)
Eq. 7.34
(69)
B 2 dτ
all
2π
Z
The magnetic field outside of the solenoid is zero and the field inside is given by,
~ = µo nI ẑ
B
(70)
where this has been used many times, but can be found in example 5.9 of Griffiths if you
wish to review.
Now we need only take the integral over the volume of the solenoid because the magnetic
field is zero elsewhere. Even better, since the magnetic field inside the solenoid is constant
we don’t need to do any integration at all. The total energy is the square of the magnetic
field multiplied by the volume of the solenoid. This should make sense in the respect that
the quantity B 2 /2µo is referred to as an energy density in both the text and lecture. For
uniform densities it is easy to find the total amount of whatever we are interested in.
W
=
1
(µo nI)2 (πR2 l)
2µo
=
1
µo πlR2 n2 I 2
2
(71)
(d) Finally, use,
1
W =
2µo
Z
2
I
B dτ −
V
~
~
(A × B) · d~a
Eq. 7.33
(72)
S
The limits of the first integral are determined from the problem statement. The upper
limit on this integral must be the radius of the solenoid because there is no magnetic field
beyond this radius. For the second integral it would at first appear as though there are
four surfaces to integrate over. There is an inner surface (whose normal vector is directed
in the −ŝ) and an outer surface (directed in the ŝ) in addition to the ends that complete the
closure of the volume.
The tube over which the integral is to be taken includes space inside and outside the
solenoid. The magnetic field outside the solenoid is zero, so that part of the surface integral in (72) that lies outside the solenoid is also zero. The surfaces at the ends of the tube
do not contribute to the integral because there the d~a points in the ẑ, while the cross prod~ ×B
~ is going to be in the ±ŝ direction depending on which end we are considering.
uct of A
The dot product in the integrand for these end surfaces is always zero.
The surface integral is therefore taken over the cylindrical surface located at s = a as this
is the only non-zero integrand.
The energy may then be calculated as,
1
W =
2µo
=
1
2µo
=
1
2µo
Z
R
Z
2π
Z
l
Z
2π
Z l
µo nI
(µo nI) · s ds dφ dz −
a φ̂ × µo nI ẑ
2
a
0
0
0
0
Z 2π Z l s2 R
1 2 2 2 2
2 2 2
µo n I l(2π) |a +
µ n I a ŝ · dφ dz ŝ
2
2 o
0
0
Z Z
1 2 2 2 2 2π l
2
2
2 2 2
πµo n I l(R − a ) + µo n I a
dφ dz
2
0
0
2
10
· (−a dφ dz ŝ)
(73)
(74)
a2
1
2 2 2
2
2
(µ n I ) πl(R − a ) + (2π)l
=
2µo o
2
(75)
=
1
(µo n2 I 2 πl)(R2 − a2 + a2 )
2
(76)
=
1
µo πlR2 n2 I 2
2
(77)
All of the different methods for determining the energy in this system return the same final
value, as should be expected.
[6.] Problem 7.29 from Griffiths
Reference figure 7.40 in Griffiths. Assume that the battery in the figure has been connected
to the LR circuit for a long time. At time t = 0 a switch is closed and this results in the
battery being disconnected from the system.
(a) Find the current in the resultant circuit, I = I(t).
(b) Find the total energy delivered to the resistor.
(c) Show that the answer in part (b) is equal to the energy initially stored in the inductor.
Solution
(a) When the battery is disconnected the entire circuit is described in terms of the inductor
and resistor alone.
dI
(78)
−L = IR
dt
Solve this differential equation,
dI
dt
R
= − I
L
(79)
R
I = A exp − t
L
(80)
The constant A is determined from the initial conditions. At time t = 0 there is a current
flowing in the circuit due to the steady state voltage supplied by the battery. This means
that I(t = 0) = Eo /R.
I(t = 0)
=
Eo
R
= A exp − · 0
R
L
A =
Eo
R
Eo
R
I(t) =
exp − t
R
L
11
(81)
(82)
(83)
Note that as t → ∞ this current goes to zero. This is critical for getting a realistic solution
to part (b).
(b) The total energy delivered to the resistor is given by,
Z
Wres = I 2 R dt
(84)
which takes advantage of the fact that the power dissipated in a resistor is P = I 2 R.
We have the current as found in part (a). The total energy delivered to the resistor must be
the power that is dissipated in the resistor over all time.
Z
Wres
=
0
∞
∞
Eo2
2R
I R dτ =
exp − t R dt
R2
L
0
∞
Eo2
L
2R
=
−
exp − t
R
2R
L 0
Z
2
= −
=
Eo2 L
[exp(−∞) − exp(0)]
2R2
Eo2 L
2R2
(85)
(86)
(87)
(88)
It is important that the current found in part (a) go to zero as time goes to infinity, otherwise, the power dissipated in the resistor would be infinite.
(c) The energy originally stored in the inductor is given by (60). In this expression the
current is the steady state value. As stated on page 317 of Griffiths, equation (60) represents
the energy stored in an inductor of value L as the current is brought up from zero to a final
value I.
This final current was found in part (b) as I = Eo /R. Therefore, the energy initially stored
in the inductor is,
1 E2
(89)
W = L o2
2 R
which matches the energy delivered to the resistor in part (b).
[7.] Problem 7.33 from Griffiths
Use the answer to problem 7.16 to solve the problems here.
a
µo Io ω
~
E(s, t) =
sin(ωt) ln
ẑ
2π
s
(90)
(a) Determine the displacement current density.
(b) Integrate the answer of part (a) to get the total displacement current. Use
Z
Id = J~d · d~a
12
(91)
(c) What is the ratio of Id to I? Letting the diameter of the outer cylinder be 2 mm, what is
the frequency at which Id is 1% of I?
Solution
(a) The displacement current density is found from the time dependent electric field using,
~
∂E
J~d ≡ o
∂t
Eq. 7.37
(92)
The only time dependence in the electric field is in the sine term, so the displacement
current density is found to be,
µo o ω 2 Io a J~d =
ln
cos(ωt) ẑ
2π
s
(93)
(b) To find the total displacement current we must integrate over the volume of the coaxial
cable from problem 7.16. Assume that the inner wire has a radius r, which will serve as a
placeholder in the calculation. At the end of the mathematical steps we will set r = 0 to get
the final answer. This is justified because we know that the inner wire is very small.
a
Z
Z
2π
J~d · s ds dφ ẑ
Id =
r
a
2π
µo o ω 2 Io a ln
cos(ωt) s ds dφ
2π
s
r
0
Z a
a
µo o ω 2 Io
(2π) cos(ωt)
s ln
ds
=
2π
s
r
Z
(94)
0
Z
=
(95)
(96)
The remaining integral may be solved using integration by parts (you are free to use any
method),
Z
a
s ln
r
a
s
a
Z 2 s2 a s
ds
ds =
ln
−
· −
2
s
2
s
r
2
Z
a
a
r2 a =
ln(1) − ln
+
s ds
2
2
r
r
r2 a a2 − r2
= − ln
+
2
r
4
= −
r2
a2 − r 2
(ln(a) − ln(r)) +
2
4
(97)
(98)
(99)
(100)
Now take r → 0 to simplify the result above. Technically, ln(0) is not defined, but we allow
(as given in Discussion on 23-4-2004) this to equal zero for this problem.
The integral in (100) becomes a2 /4.
13
Completing the solution for the displacement current,
Id =
=
a2
µo o ω 2 Io
(2π) cos(ωt)
2π
4
(101)
1
µo o ω 2 a2 Io cos(ωt)
4
(102)
(c) Recall from problem 7.16 that I = Io cos(ωt). The ratio is,
Id
I
=
1
µ ω 2 a2 Io
4 o o
cos(ωt)
Io cos(ωt)
=
µ o o ω 2 a2
4
(103)
To calculate the frequency at which this ratio is 1% is found more easily by substituting
µo o = 1/c2 , where c is the speed of light.
Id
I
=
0.01 =
ω2 =
=
ω 2 a2
4c2
(104)
0.04c2
a2
(105)
0.04(3.0 × 108 )2
0.0012
(106)
= 3.6 × 1021
(107)
ω = 6.0 × 1010
(108)
The frequency we want is given by f = ω/2π.
f
=
ω
2π
=
9.5 × 109 Hz
(109)
This frequency is considerably higher than those used in experiments during the time of
Faraday. This is the low end of the microwave region of the electromagnetic spectrum.
[8.] Problem 7.53 from Griffiths
Two coils are wrapped around a cylinder (that may be neglected for all purposes beyond
providing a form for the coil winding) such that the same magnetic flux passes through
both sets of coils. Label coils 1, with N1 turns, and 2, with N2 turns, as the primary and
secondary, respectively. Letting the current in the primary change in time, show that
E2
N2
=
E1
N1
(110)
where E2 is the emf induced in coil 2 by the primary, and E1 is the back emf in coil 1.
Reference figure 7.54 in Griffiths.
14
Solution
These coils may be treated as solenoids. As such, the magnetic field of the primary coil
is known. Also, the magnetic flux through the secondary is easily found since the field is
uniform inside the solenoids. Each turn of the secondary contributes to the final emf, so
there is a factor of N2 in the calculation.
Let the radius of the solenoids be r, the fact that this is not given in the problem implies
that it will not matter in the final solution.
The emf of coil 2 is,
E2
−N2
=
d
dΦ
= −N2 B1 · πr2
dt
dt
= −N2
d
µo n1 Iπr2
dt
= −N2 µo n1 πr2
dI
dt
(111)
(112)
(113)
The back emf of coil 1 is found using (37). The self-inductance of a solenoid is given in (61).
Putting all of this together we have the back emf of the primary,
E1
=
−L1
dI
dt
= −µo πr2 n21 l
dI
dt
= −N1 µo n1 πr2
dI
dt
(114)
(115)
Note that in this calculation the self-inductance already accounts for the total number of
turns so I did not multiply by N1 . The factor of N1 is taken out of the expression for L from
n1 = N1 /l, where l is the length of the solenoid.
Now that we have both expressions for the emf we may write their ratio,
E2
E1
=
−N2 µo n1 πr2 dI
dt
−N1 µo n1 πr2 dI
dt
15
=
N2
N1
(116)