Download PHYS 110B - HW #3

PHYS 110B - HW #3
Fall 2005, Solutions by David Pace
Equations referenced as, ”Eq. #” 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) I’ll reference the smaller loop as #1 and the larger loop as #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,
B(z) =
µo I
R2
2 (R2 + z 2 )3/2
Eq. 5.38
(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. Recall that there is a form of the dipole magnetic field that has a specific
spherical coordinate representation,
~ dip = µo m 2 cos θ r̂ + sin θ θ̂
B
Eq. 5.86
(5)
4πr3
where m is the magnetic moment of the small loop. Recall from Eq. 5.84 that m = Iπa2 .
1
Now we must define a surface that has the ring of radius b as its perimeter. The problem will be
simplified if we choose this surface such that its area normal is parallel to the magnetic field in (5).
~ One such area is the semi-spherical area traced
In other words, choose an area such that d~a k B.
out by a vector from the center of loop 1 that terminates on the wire of loop 2. The differential area
vector of this area is directed along r̂, which greatly simplifies the math in (4).
Figure 1 shows the geometry of this method. Defining a spherical coordinate system with the
smaller loop at the center it is seen that a surface with, d~a = r2 sin θ dθ dφ r̂, exists by sweeping
the vector, ~r, from the center of 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.
p
r =
b2 + z 2
(8)
sin θ0 =
b
r
2
=
√
b2
b
+ z2
(9)
Continue solving for the flux through loop 2,
θ0
µo Ia2
(cos θ) sin θ dθ
2r
0
Z θ0
µ πIa2
√o
(cos θ) sin θ dθ
b2 + z 2 0
θ 0
1
µo πIa2
2
√
sin θ
b2 + z 2 2
0
b2
µ πIa2
√o
2
2 b2 + z 2 b + z 2
Z
Φ2 = 2π
(10)
=
(11)
=
=
=
µo πa2 b2 I
2(b2 + z 2 )3/2
(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 specific loop. In this problem the currents through the wires are the
same, i.e. I1 = I2 .
Finding M12 ,
M12
=
Φ1
I
=
µo πa2 b2
2(b2 + z 2 )3/2
(16)
M21
=
Φ2
I
=
µo πa2 b2
2(b2 + z 2 )3/2
(17)
For M21 ,
From (16) and (17) we confirm that the mutual inductances are equivalent. This serves as a demonstration that mutual inductance is a system property. The system consists of two wire loops and has
a generic inductance. Since this is a property of the entire system it should not matter which loop
we consider in order to solve for it. With this in mind, problems like this one can usually be solved
much more easily by choosing the simpler mathematical path. The method of part (a) is the simpler
method in this example.
[2.] Carter Problem
The two infinitely long antiparallel currents and the rectangle in the figure below all lie in the same
plane. A time changing current, I(t) = Io (1−exp(−t/τ )), is driven in the rectangular loop. Calculate
the EMF generated in the circuit composed of the antiparallel currents (the circuit is actually a very
long rectangular loop, with the two other sides out at infinity).
Solution
The emf generated in the circuit is given by,
E
= −
dΦ
dt
3
Eq. 7.17
(18)
One method to solve this problem involves determining the magnetic field generated by the square
circuit, using that to calculate the magnetic flux through the anti-parallel currents, and then taking
the time derivative to solve for the emf.
The problem with this method is that the field generated by the square circuit is very difficult to
determine. Even if we worked it out, using that field to solve for the total flux through the antiparallel currents will be even more difficult (far away from the square, its field might look dipolelike, which does not suggest any straightforward method for determining this flux).
Mutual inductance is symmetric, however, meaning that we can place our current, I(t), in the antiparallel circuit and solve for the magnetic flux through the square. This flux will be equivalent to
that through the anti-parallel currents when I(t) flows through the square.
Putting I(t) in the anti-parallel current circuit, we determine the flux through the square due to two
wires distances d and D + d away (see the figure on the homework assignment sheet). Since these
anti-parallel currents are actually part of a complete circuit, even the time dependent current will
be flowing anti-parallel. This means that they will partially cancel out in the area of the square. The
magnetic field in the region of the square is (using cylindrical coordinates centered on the left-most
wire),
µo I(t)
µo I(t)
−
2πs
2π(s − D)
µo Io
t
1
1
1 − exp −
−
2π
τ
s s−D
µo Io
t
D
1 − exp −
2π
τ
s(s − D)
B =
=
=
Now solve for the flux through the square (d~a k φ̂, into/out of the page),
Z
~ · d~a
Φ =
B
=
=
(20)
(21)
(22)
D+d+a Z b
µo Io
t
D
1 − exp −
ds dz
2π
τ
s(s − D)
D+d
0
Z D+d+a
ds
µo Io bD
t
1 − exp −
2π
τ
s(s − D)
D+d
Z
(19)
(23)
(24)
At this point a useful integral is,
Z
dx
x(ax + c)
=
1 x ln
c ax + c (25)
The flux is (with a → 1 and c → −D),
Φ
µo Io bD
t
1 s D+d+a
=
1 − exp −
− ln 2π
τ
D
s − D D+d
d(D + d + a) µo Io b
t
= −
1 − exp −
ln 2π
τ
(D + d)(d + a) (26)
(27)
This is also the flux that will pass through the anti-parallel currents when the current, I(t), flows in
the square.
4
The emf induced in the anti-parallel currents is,
d(D + d + a) µo Io b
t
d
−
1 − exp −
ln E = −
dt
2π
τ
(D + d)(d + a) t
µo Io b d(D + d + a) exp −
=
ln 2πτ
(D + d)(d + a) τ
(28)
(29)
Notice that the induced emf goes to zero as t → ∞. The current in the square goes to a constant
value as t → ∞. When this current is at a steady state value the flux through the anti-parallel
currents is constant. This constant flux induces no emf, as (29) shows.
[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
Φ =
=
=
h
µo I
ds dz
a
0 2πs
b
µo hIo cos(ωt)
ln
2π
a
(4π × 10−7 )(0.01)(0.5) cos(ωt)
ln(2)
2π
= 6.93 × 10−10 cos(120πt)
(30)
(31)
(32)
(33)
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)
(34)
The emf in the toroid is found using,
E
= −
dΦ
dt
(35)
E
= −
d
6.93 × 10−7 cos(120πt)
dt
(36)
= −6.93 × 10−7 (−120π) sin(120πt)
(37)
= 2.61 × 10−4 sin(120πt)
(38)
where emf is in units of Volts.
The current in the resistor is,
I
=
E
R
5.22 × 10−7 sin(120πt)
=
(39)
in units of Amperes.
(b) The back emf, Eb , of a system is that due to its self-inductance.
dI
dt
Eb = −L
Eq. 7.26
(40)
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
(41)
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)
(42)
(43)
(44)
and the sign is inconsequential here. You may represent this as an absolute value if you wish. We
know that conceptually this emf must be opposing the ”applied” emf due to the free current that
we are running in the axial wire.
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
(45)
The back emf is approximately 1000 times smaller than the emf directly induced by the center wire.
6
[4.] 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.
(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.
Solution
(a) We are supposed to use,
1
W = LI 2
2
Eq. 7.29
(46)
The current is given in the problem. From the lecture of 10/14/2005 we know the self-inductance
of this solenoid is,
L = µo πlR2 n2
(47)
where this is also found in Griffiths’ problem 7.22 (essentially what the lecture covered).
Therefore, the energy stored in this system is,
1
W = µo πlR2 n2 I 2
2
(48)
(b) The equation to use is,
W =
1
2
I
~ · I)
~ dl
(A
Eq. 7.30
(49)
From example 5.12 (page 238) the vector potential is,
~ =
A
µo nI
s φ̂
2
~ =
A
µo nI R2
φ̂
2 s
s<R
s>R
(50)
(51)
where it should be noted that the example is solving for an infinite solenoid. This is still applicable
here because the problem states this is a long solenoid and we are including the length dependence
in our final answers.
The integrand in (49) 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 (49) 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.
7
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,
I
Z
nl 2π µo nI
nl
~
~
(A · I) dl =
(52)
W =
R · I(Rdφ)
2
2 0
2
(c) Now use,
W =
1
2µo
Z
=
nl µo nI
RI(R2π)
2 2
(53)
=
1
µo πlR2 n2 I 2
2
(54)
Eq. 7.34
(55)
B 2 dτ
all
The magnetic field outside of the solenoid is zero and the field inside is given by,
~ = µo nI ẑ
B
(56)
where this has been used many times and can be found in example 5.9 of Griffiths if you wish to
review.
Now we need to perform the integral only through 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 we do not need
to bother with the formal integration.
W
=
1
(µo nI)2 (πR2 l)
2µo
=
1
µo πlR2 n2 I 2
2
(57)
(d) Finally, use,
1
W =
2µo
Z
2
I
B dτ −
V
~
~
(A × B) · d~a
Eq. 7.33
(58)
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 (58) that lies outside
the solenoid is also zero. The surfaces at the ends of the tube do not contribute to the integral because
~×B
~ is going to be in the ±ŝ
there the d~a points in the ±ẑ direction, while the cross product of A
direction depending on which end we are considering. 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.
8
The energy may then be calculated as,
W
=
1
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
0
0
a
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 2π Z l
1
πµ2o n2 I 2 l(R2 − a2 ) + µ2o n2 I 2 a2
dφ dz
2
0
0
W
2
· (−a dφ dz ŝ)
(59)
(60)
=
1
a2
2 2 2
2
2
(µ n I ) πl(R − a ) + (2π)l
2µo o
2
(61)
=
1
(µo n2 I 2 πl)(R2 − a2 + a2 )
2
(62)
=
1
µo πlR2 n2 I 2
2
(63)
All of the different methods for determining the energy in this system return the same final value,
as should be expected.
[5.] 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
−L
= IR
(64)
dt
where I have put the emf sources on the left side and the emf ”sinks” (dissipation elements) on the
right side. See footnote 11 of example 7.12 in Griffiths for an additional explanation.
Solve this differential equation,
dI
dt
R
= − I
L
(65)
R
I = A exp − t
L
(66)
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.
9
I(t = 0)
=
Eo
R
R
= A exp − · 0
L
A =
Eo
R
Eo
R
I(t) =
exp − t
R
L
(67)
(68)
(69)
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
(70)
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
∞
2
I R dτ
=
=
0
=
= −
=
∞
Eo2
2R
exp − t R dt
R2
L
0
Eo2
L
2R ∞
−
exp − t
R
2R
L 0
Z
Eo2 L
[exp(−∞) − exp(0)]
2R2
Eo2 L
2R2
(71)
(72)
(73)
(74)
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 (46). In this expression the current is the
steady state value. As stated on page 317 of Griffiths, equation (46) 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
W = L o2
(75)
2 R
which matches the energy delivered to the resistor in part (b).
[6.] Carter Problem
A toroidal coil of N turns has a major radius of b and a square cross-section of side a. If a current I
flows in the coil, calculate the stored magnetic energy two ways:
10
(1) Using W = LI 2 /2
(2) By integrating the magnetic field energy density (over all space).
Solution
Figure 2 illustrates the geometry of this problem.
Figure 2: The toroid of problem [6.]. The major radius is b, meaning that the inner radius is
equal to b − a/2 and the outer radius is equal to b + a/2. There are N turns, of which only a
few have been drawn.
(1) The problem gives the current, I. Example 7.11 in Griffiths solves for the self-inductance of a
rectangular toroid. Following that example and substituting the inner and outer radii as described
in figure 2, the self-inductance of this toroid is,
b + a2
µo aN 2
ln
(76)
L =
2π
b − a2
Using W = LI 2 /2, the stored magnetic energy in the toroid is,
b + a2
µo aN 2 I 2
W =
ln
4π
b − a2
(77)
(2) Now we want to use (55) to determine the stored magnetic energy. Griffiths has provided an
expression for the magnetic field inside a toroid,
~ =
B
µo N I
φ̂
2πs
Eq. 5.58
(78)
where φ̂ is directed into and out of the page, and the magnetic field is zero outside of the toroid.
11
Integrating over all space is simplified since the magnetic field is zero outside of the solenoid.
forming the integration throughout the volume of the toroid leads to,
Z b+a/2 Z 2π Z a µo N I 2
1
W =
s ds dφ dz
2µo b−a/2 0
2πs
0
Z b+a/2
ds
1 µ2o N 2 I 2
(2πa)
=
2
2µo
4π
b−a/2 s
b + a2
µo aN 2 I 2
=
ln
4π
b − a2
Per-
(79)
(80)
(81)
where the factor of 2πa in (80) comes from the integrations over φ and z.
The results of (77) and (81) are equivalent, as expected.
[7.] Problem 7.33 from Griffiths
Use the answer to problem 7.16 to solve the problems here.
~ t) = µo Io ω sin(ωt) ln a ẑ
E(s,
2π
s
(82)
(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
(83)
(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
(84)
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
(85)
2π
s
(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.
Z a Z 2π
Id =
(86)
J~d · s ds dφ ẑ
r
=
=
0
a Z 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
12
(87)
(88)
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
(89)
(90)
= −
r2 a a2 − r2
ln
+
2
r
4
(91)
= −
r2
a2 − r2
(ln(a) − ln(r)) +
2
4
(92)
Now take r → 0 to simplify the result above. Technically, the term ln(0) is not defined, but we allow
it to equal zero for this problem. The physical argument is that real wires cannot have a radius of
zero. As the radius gets very small, the r2 dominates and the entire first term in (92) goes to zero.
The integral in (92) becomes a2 /4.
Completing the solution for the displacement current,
Id =
=
µo o ω 2 Io
a2
(2π) cos(ωt)
2π
4
(93)
1
µo o ω 2 a2 Io cos(ωt)
4
(94)
(c) Recall from problem 7.16 that I = Io cos(ωt). The ratio is,
Id
I
1
2 2
4 µo o ω a Io cos(ωt)
=
Io cos(ωt)
=
µo o ω 2 a2
4
(95)
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
(96)
0.04c2
a2
(97)
0.04(3.0 × 108 )2
0.0012
(98)
= 3.6 × 1021
(99)
ω = 6.0 × 1010
(100)
The frequency we want is given by f = ω/2π.
f
=
ω
2π
=
13
9.5 × 109 Hz
(101)
This frequency is considerably higher than those used in experiments during the time of Faraday.
Such a frequency is in the low end of the microwave region of the electromagnetic spectrum.
[8.] Problem 7.54 from Griffiths
Use problem 7.53 as a reference and explanation of a transformer.
In a transformer, the ratio of the input and output voltages is given by,
V2
V1
=
N2
N1
(102)
which appears to violate energy conservation when N2 > N1 . Energy is conserved, however, because even though the output voltage is increased in this situation, the output current decreases.
Power is given by, P = IV . This problem aims to further demonstrate this behavior.
(a) In the ideal case, the same flux passes through all the turns of the primary and secondary coils.
Using this fact, show that M 2 = L1 L2 , where M is the mutual inductance of this transformer system
and the L’s represent the self-inductances of each coil.
(b) Drive the primary coil with Vin = V1 cos(ωt), and connect a resistor of value R to the secondary.
Show that the currents satisfy,
L1
dI1
dI2
+M
dt
dt
= V1 cos(ωt)
(103)
L2
dI2
dI1
+M
dt
dt
= −I2 R
(104)
(c) Use parts (a) and (b) to solve for the currents. There is no DC component to I1 .
(d) Show that the ratio of output voltage (Vout = V2 = I2 R) to input voltage (Vin = V1 = V1 cos(ωt))
satisfies (102).
(e) Determine the input and output powers and show that they are equal over one complete cycle.
Solution
(a) The magnetic flux through the primary (#1) coil is related to the mutual inductance of the system by (15). The flux in that expression represents the entire amount of flux throughout the entire
solenoid that represents the primary. Rewrite this as,
N1 Φ1 = M12 I2
(105)
where the left side represents the total flux passing through the primary (Φ1 represents the flux
passing through only one turn) and I2 is the current flowing in the secondary.
Considering the current flowing in the primary allows us to write a similar expression regarding
the self-inductance,
N1 Φ1 = L1 I1
14
(106)
The equivalent expressions for the secondary coil are,
N2 Φ2 = M21 I1
(107)
N2 Φ2 = L2 I2
(108)
Pairing (105) and (106) to solve for the self-inductance of the primary gives,
L1 =
M12 I2
I1
(109)
M21 I1
I2
(110)
Once again, the secondary has a similar result,
L2 =
Multiplying these self-inductances,
L1 L2 =
M12 I2 M21 I1
I1
I2
(111)
= M12 M21
(112)
= M2
(113)
where in (112) I have made use of the symmetry property of mutual inductance (Eq. 7.23: M21 =
M12 ).
(b) Write out the voltage balance equation (i.e. apply Kirchoff’s law) in the following form,
Sources = Sinks
(114)
For the primary coil there are no sinks (dissipative voltage elements). Writing out all of the voltage
sources gives,
V1 cos(ωt) + Eb,L + Eb,M
= 0
(115)
dI1
dI2
−M
dt
dt
= 0
(116)
V1 cos(ωt) − L1
dI1
dI2
(117)
+M
dt
dt
where the Eb terms represent the back emf’s contributed by the inductive elements. These are due to
the back emf of the primary’s self-inductance and the back emf of the interaction between the coils.
The relations between voltage and inductance are given by (18) with the appropriate substitution
for Φ.
V1 cos(ωt) = L1
The secondary coil does have a dissipative element. In this case,
Eb,L + Eb,M
= I2 R
(118)
−L2
dI2
dI1
−M
dt
dt
= I2 R
(119)
L2
dI2
dI1
+M
dt
dt
= −I2 R
(120)
15
(c) At this point we have two equations and two unknowns. The following is one method for solving
for the currents. Multiply (117) by L2 and multiply (120) by M .
L2 V1 cos(ωt) = L2 L1
−M I2 R = M L2
dI1
dI2
+ L2 M
dt
dt
(121)
dI2
dI1
+ M2
dt
dt
(122)
Subtract (121) from (122),
−L2 V1 cos(ωt) − M I2 R = 0
(123)
where the result in (a) has been applied.
The current, I2 , is,
I2 = −
L2 V1
cos(ωt)
MR
(124)
Substitute (124) into (117) to solve for I1 .
dI1
d
L2 V1
V1 cos(ωt) = L1
+M
−
cos(ωt)
dt
dt
MR
dI1 L2 V1 ω
+
sin(ωt)
dt
R
1
L2 V1 ω
V1 cos(ωt) −
sin(ωt)
L1
R
Z
1
L2 V1 ω
sin(ωt) dt
V1 cos(ωt) −
L1
R
1
L2 V1 ω −1
1
cos(ωt)
V1 sin(ωt) −
L1
ω
R
ω
V1 sin(ωt) L2 cos(ωt)
+
L1
ω
R
V1 cos(ωt) = L1
dI1
dt
=
Z
dI1 =
I1 =
=
(125)
(126)
(127)
(128)
(129)
(130)
(d) The input voltage is given in (b), and the output voltage can be calculated since we have solved
for I2 .
Vout = I2 R
=
L2 V1
cos(ωt)
M
(131)
(132)
where I have removed the negative sign from I2 because at this point we are interested in the absolute value of the ratio.
16
The ratio is,
Vout
Vin
=
L2 V1
1
cos(ωt)
M
V1 cos(ωt)
(133)
=
L2
M
(134)
Use (108) to substitute for L2 , and (105) to substitute for M (which represents the mutual inductance
of the system).
Vout
N 2 Φ2
I2
=
(135)
Vin
I2
N 1 Φ1
=
N 2 Φ2
N 1 Φ1
(136)
=
N2
N1
(137)
where the last step is possible because the fluxes through each turn are equal (recalling that in this
formulation Φ1 represents the magnetic flux through only one turn).
(e) The input power is,
(138)
Pin = Vin I1
V1 sin(ωt) L2 cos(ωt)
= V1 cos(ωt)
+
L1
ω
R
For a sinusoidal function, f , the average over one complete cycle is given by,
Z 2π
1
hf i =
f dt
2π 0
The average power over one cycle is,
Z 2π
1
V12 sin(ωt) L2 cos(ωt)
hPin i =
cos(ωt)
+
dt
2π 0
L1
ω
R
Z 2π V12
1
L2 cos2 (ωt)
cos(ωt) sin(ωt) +
dt
=
2πL1 0
ω
R
V12
L2
=
0+
2πL1
2R
=
L2 V12
4πRL1
(139)
(140)
(141)
(142)
(143)
(144)
Now consider the power in the output circuit,
Pout = Vout I2
=
2
1 L2 V1
cos(ωt)
R M
17
(145)
(146)
The average over one cycle,
hPout i =
=
2
1 L2 V1
cos(ωt) dt
R M
0
1
L22 V12
2
2πRM
2
1
2π
Z
2π
(147)
(148)
=
L22 V12
4πRL1 L2
(149)
=
L2 V12
4πRL1
(150)
The powers are equal and energy is conserved.
18