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Inductance and Magnetic Energy
- Chapter
- 11.1 Mutual Inductance 11-
- Example 11.1 Mutual Inductance of Two Concentric Coplanar Loops 11-
- 11.2 Self-Inductance 11-
- Example 11.2 Self-Inductance of a Solenoid........................................................ 11-
- Example 11.3 Self-Inductance of a Toroid 11-
- Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid 11-
- 11.3 Energy Stored in Magnetic Fields 11-
- Example 11.5 Energy Stored in a Solenoid 11-
- Animation 11.1: Creating and Destroying Magnetic Energy............................ 11-
- Animation 11.2: Magnets and Conducting Rings 11-
- 11.4 RL Circuits 11-
- 11.4.1 Self-Inductance and the Modified Kirchhoff's Loop Rule....................... 11-
- 11.4.2 Rising Current.......................................................................................... 11-
- 11.4.3 Decaying Current 11-
- 11.5 LC Oscillations 11-
- 11.6 The RLC Series Circuit 11-
- 11.7 Summary......................................................................................................... 11-
- 11.8 Appendix 1: General Solutions for the RLC Series Circuit 11-
- 11.8.1 Quality Factor 11-
- 11.9 Appendix 2: Stresses Transmitted by Magnetic Fields 11-
- Animation 11.3: A Charged Particle in a Time-Varying Magnetic Field 11-
- 11.10 Problem-Solving Strategies 11-
- 11.10.1 Calculating Self-Inductance................................................................... 11-
- 11.10.2 Circuits containing inductors 11-
- 11.11 Solved Problems 11-
- 11.11.1 Energy stored in a toroid........................................................................ 11-
- 11.11.2 Magnetic Energy Density 11-
- 11.11.3 Mutual Inductance 11-
- 11.11.4 RL Circuit............................................................................................... 11-
- 11.11.5 RL Circuit............................................................................................... 11-
- 11.11.6 LC Circuit............................................................................................... 11-
- 11.12 Conceptual Questions 11-
- 11.13 Additional Problems 11-
- 11.13.1 Solenoid 11-
- 11.13.2 Self-Inductance 11-
- 11.13.3 Coupled Inductors.................................................................................. 11-
- 11.13.4 RL Circuit............................................................................................... 11-
- 11.13.5 RL Circuit............................................................................................... 11-
- 11.13.6 Inductance of a Solenoid With and Without Iron Core 11-
- 11.13.7 RLC Circuit 11-
- 11.13.8 Spinning Cylinder 11-
- 11.13.9 Spinning Loop........................................................................................ 11-
1 henry = 1 H = 1 T m /A⋅^2 (11.1.4)
We shall see that the mutual inductance M (^) 21 depends only on the geometrical properties
of the two coils such as the number of turns and the radii of the two coils.
In a similar manner, suppose instead there is a current I 2 in the second coil and it is varying with time (Figure 11.1.2). Then the induced emf in coil 1 becomes
12 12 1 2 1 coil 1
d d N dt dt
ε d
= − = − ∫∫ B ⋅ A
G G
and a current is induced in coil 1.
Figure 11.1.2 Changing current in coil 2 produces changing magnetic flux in coil 1.
This changing flux in coil 1 is proportional to the changing current in coil 2,
12 2 1 12
d N M dt dt
Φ dI = (11.1.6)
where the proportionality constant M 12 is another mutual inductance and can be written
as
1 12 12 2
N
M
I
However, using the reciprocity theorem which combines Ampere’s law and the Biot- Savart law, one may show that the constants are equal:
M 12 (^) = M (^) 21 ≡ M (11.1.8)
Example 11.1 Mutual Inductance of Two Concentric Coplanar Loops
Consider two single-turn co-planar, concentric coils of radii R 1 and R 2 , with R 1 (^) R 2 , as
shown in Figure 11.1.3. What is the mutual inductance between the two loops?
Figure 11.1.3 Two concentric current loop
Solution:
The mutual inductance can be computed as follows. Using Eq. (9.1.15) of Chapter 9, we see that the magnetic field at the center of the ring due to I 1 in the outer coil is given by
0 1 1 (^21)
I
B
R
Since R 1 (^) R 2 , we approximate the magnetic field through the entire inner coil by B 1.
Hence, the flux through the second (inner) coil is
2 0 1 2 0 1 2 21 1 2 2 (^2 1 )
I I R
B A R
R R
Thus, the mutual inductance is given by
2 21 0 2 1 2 1
R
M
I R
The result shows that M depends only on the geometrical factors, R 1 and R 2 , and is
independent of the current I 1 in the coil.
11.2 Self-Inductance
Consider again a coil consisting of N turns and carrying current I in the counterclockwise direction, as shown in Figure 11.2.1. If the current is steady, then the magnetic flux through the loop will remain constant. However, suppose the current I changes with time,
Figure 11.2.2 Solenoid
Solution:
Ignoring edge effects and applying Ampere’s law, the magnetic field inside a solenoid is given by Eq. (9.4.3):
0 ˆ 0
NI
nI l
B = μ^ k =μ k^ ˆ
G
where is the number of turns per unit length. The magnetic flux through each turn is
n = N / l
2 B 0 (^ ) 0 Φ = BA = μ nI ⋅ π R = μ nI π R^2 (11.2.5)
Thus, the self-inductance is
2 2 0 L N^ B n I
μ π R l
We see that L depends only on the geometrical factors ( n , R and l ) and is independent of the current I.
Example 11.3 Self-Inductance of a Toroid
Calculate the self-inductance of a toroid which consists of N turns and has a rectangular cross section, with inner radius a , outer radius b and height h , as shown in Figure 11.2.3(a).
(a) (b)
Figure 11.2.3 A toroid with N turns
Solution:
According to Ampere’s law discussed in Section 9.3, the magnetic field is given by
∫ B^ ⋅^ d^ s =^ ∫ Bds^ =^ B^ ∫ ds^ = B^ (2^^ π^ r^ )=μ^0 N^ (11.2.7)
G G
v v v I
or
0 2
NI
B
r
μ π
The magnetic flux through one turn of the toroid may be obtained by integrating over the rectangular cross section, with dA = h dr as the differential area element (Figure
11.2.3b):
(^0 0) ln 2 2
b B (^) a
NI NIh b d hdr r a
μ μ π π
∫∫ B^ A ∫
G G
The total flux is N Φ B. Therefore, the self-inductance is
2 (^0) ln 2
L N^ B^ N h b I a
Again, the self-inductance L depends only on the geometrical factors. Let’s consider the situation where a b. In this limit, the logarithmic term in the equation above may be expanded as
− a
ln ln 1
b b a b a a
a a
⎛ ⎞ ⎛ −^ ⎞ −
⎜ ⎟ =^ ⎜ +^ ⎟≈
and the self-inductance becomes
2 2 0 0 2 2
N h b a N A N A L a a l
μ μ π π
2 μ 0
) a
where A = h b ( − a is the cross-sectional area, and l = 2 π. We see that the self-
inductance of the toroid in this limit has the same form as that of a solenoid.
Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid
A long solenoid with length l and a cross-sectional area A consists of N 1 turns of wire. An insulated coil of N 2 turns is wrapped around it, as shown in Figure 11.2.4.
where k is the “coupling coefficient.” In our example, we have k = 1 which means that all of the magnetic flux produced by the solenoid passes through the outer coil, and vice versa, in this idealization.
11.3 Energy Stored in Magnetic Fields
Since an inductor in a circuit serves to oppose any change in the current through it, work must be done by an external source such as a battery in order to establish a current in the inductor. From the work-energy theorem, we conclude that energy can be stored in an inductor. The role played by an inductor in the magnetic case is analogous to that of a capacitor in the electric case.
The power, or rate at which an external emf ε (^) extworks to overcome the self-induced emf
ε (^) L and pass current I in the inductor is
ext L ext
dW P dt
= = I ε (11.3.1)
If only the external emf and the inductor are present, then ε (^) ext = −ε (^) L which implies
ext L L
dW dI P I dt dt
= = − ε = + IL
If the current is increasing with , then which means that the external source is doing positive work to transfer energy to the inductor. Thus, the internal energy of the inductor is increased. On the other hand, if the current is decreasing with , we then have. In this case, the external source takes energy away from the inductor, causing its internal energy to go down. The total work done by the external source to increase the current form zero to
dI / dt > P > 0
U B
dI / dt < P < 0
I is then
2 ext ext (^0)
I
W = ∫ dW = ∫ LI dI = LI (11.3.3)
This is equal to the magnetic energy stored in the inductor:
U B = LI (11.3.4)
The above expression is analogous to the electric energy stored in a capacitor:
E
Q
U
C
We comment that from the energy perspective there is an important distinction between an inductor and a resistor. Whenever a current I goes through a resistor, energy flows into the resistor and dissipates in the form of heat regardless of whether I is steady or time-
dependent (recall that power dissipated in a resistor is ). On the other hand,
energy flows into an ideal inductor only when the current is varying with. The energy is not dissipated but stored there; it is released later when the current decreases with. If the current that passes through the inductor is steady, then there is no change in energy since.
2 P R = IVR = I R dI / dt > 0
dI / dt < 0 PL = LI dI ( / dt ) = 0
Example 11.5 Energy Stored in a Solenoid
A long solenoid with length l and a radius R consists of N turns of wire. A current I passes through the coil. Find the energy stored in the system.
Solution:
Using Eqs. (11.2.6) and (11.3.4), we readily obtain
2 2 2 0
U (^) B = LI = μ n I π R (^2) l (11.3.6)
The result can be expressed in terms of the magnetic field strength B = μ 0 nI :
2 2 2 2 0 0 0
B 2 2
B
U μ nI π R l π μ μ
= = R l ) (11.3.7)
Since π R^2 l is the volume within the solenoid, and the magnetic field inside is uniform, the term
2
(^20) B
B
u μ
may be identified as the magnetic energy density, or the energy per unit volume of the magnetic field. The above expression holds true even when the magnetic field is non- uniform. The result can be compared with the energy density associated with an electric field:
2 0
E 2
u = ε E (11.3.9)
same time and at the same rate, in a manner that has been pre-arranged. Once the charges on the rings start to decelerate, the magnetic field in the space between the rings starts to decrease in magnitude. As the magnetic flux through the rings decreases, Faraday’s law tells us that there is now an electric field induced by the time-changing magnetic field that is circulating counterclockwise as seen from above. The force on the charges due to this electric field is thus in the same direction as the motion of the charges. In this situation the agents have work done on them as they try to spin the charges down.
Over the course of the “destroy” animation associated with Figure 11.3.1, the strength of the magnetic field decreases, and this energy flows from the field back to the path along which the charges move, and is now being provided to the agents trying to spin down the moving charges. The energy provided to those agents as they destroy the magnetic field is exactly the amount of energy that they put into creating the magnetic field in the first place, neglecting radiative losses (such losses are small if we move the charges at speeds small compared to the speed of light). This is a totally reversible process if we neglect such losses. That is, the amount of energy the agents put into creating the magnetic field is exactly returned to the agents as the field is destroyed.
There is one final point to be made. Whenever electromagnetic energy is being created,
an electric charge is moving (or being moved) against an electric field ( ).
Whenever electromagnetic energy is being destroyed, an electric charge is moving (or
being moved) along an electric field (
q v E ⋅ < 0
G^ G
q v E ⋅ > 0
G G
). This is the same rule we saw above
when we were creating and destroying electric energy above.
Animation 11.2: Magnets and Conducting Rings
In the example of Faraday’s law that we gave above, the sense of the electric field associated with a time-changing magnetic field was always such as to try to resist change. We consider another example of Faraday’s law that illustrates this same tendency in a different way.
Figure 11.3.2 A perfectly conducting ring falls on the axis of a permanent magnet. The induced currents and the resulting magnetic field stresses are such as to slow the fall of the ring. If the ring is light enough (or the magnet strong enough), the ring will levitate above the magnet.
In Figure 11.3.2, we show a permanent magnet that is fixed at the origin with its dipole moment pointing upward. On the z -axis above the magnet, we have a co-axial, conducting, non-magnetic ring with radius a , inductance L, and resistance R. The center of the conducting ring is constrained to move along the vertical axis. The ring is released from rest and falls under gravity toward the stationary magnet. Eddy currents arise in the ring because of the changing magnetic flux and induced electric field as the ring falls toward the magnet, and the sense of these currents is to repel the ring when it is above the magnet.
This physical situation can be formulated mathematically in terms of three coupled ordinary differential equations for the position of the ring, its velocity, and the current in the ring. We consider in Figure 11.3.2 the particular situation where the resistance of the ring (which in our model can have any value) is identically zero, and the mass of the ring is small enough (or the field of the magnet is large enough) so that the ring levitates above the magnet. We let the ring begin at rest a distance 2 a above the magnet. The ring begins to fall under gravity. When the ring reaches a distance of about a above the ring, its acceleration slows because of the increasing current in the ring. As the current increases, energy is stored in the magnetic field, and when the ring comes to rest, all of the initial gravitational potential of the ring is stored in the magnetic field. That magnetic energy is then returned to the ring as it “bounces” and returns to its original position a distance 2 a above the magnet. Because there is no dissipation in the system for our particular choice of R in this example, this motion repeats indefinitely.
What are the important points to be learned from this animation? Initially, all the free energy in this situation is stored in the gravitational potential energy of the ring. As the ring begins to fall, that gravitational energy begins to appear as kinetic energy in the ring. It also begins to appear as energy stored in the magnetic field. The compressed field below the ring enables the transmission of an upward force to the moving ring as well as a downward force to the magnet. But that compression also stores energy in the magnetic field. It is plausible to argue based on the animation that the kinetic energy of the downwardly moving ring is decreasing as more and more energy is stored in the magnetostatic field, and conversely when the ring is rising.
Figure 11.3.3 shows a more realistic case in which the resistance of the ring is finite. Now energy is not conserved, and the ring eventually falls past the magnet. When it passes the magnet, the sense of the induced electric field and thus of the eddy currents reverses, and the ring is now attracted to the magnet above it, which again retards its fall.
There are many other examples of the falling ring and stationary magnet, or falling magnet and stationary ring, given in the animations at this link. All of them show that the effect of the electric field associated with a time-changing magnetic field is to try to keep things the same. In the limiting case of zero resistance, it can in fact achieve this goal, e.g. in Figure 11.3.2 the magnetic flux through the ring never changes over the course of the motion.
To show how to do this, consider the circuit shown in Figure 11.4.1. We have a battery, a resistor, a switch S that is closed at t = 0, and a “one-loop inductor.” It will become clear what the consequences of this “inductance” are as we proceed. For t > 0, current will flow in the direction shown (from the positive terminal of the battery to the negative, as usual). What is the equation that governs the behavior of our current I ( ) t for t > 0?
To investigate this, apply Faraday's law to the open surface bounded by our circuit, where
we take d A to be out of the page, and d
G
s
G
right-handed with respect to that choice
(counter-clockwise). First, what is the integral of the electric field around this circuit? There is an electric field in the battery, directed from the positive terminal to the negative terminal, and when we go through the battery in the direction of d s
G
that we have chosen,
we are moving against that electric field, so that E ⋅ d s < 0
G G
. Thus the contribution of the
battery to our integral is− ε. Then, there is an electric field in the resistor, in the direction
of the current, so when we move through the resistor in that direction, is greater
than zero, and that contribution to our integral is
E ⋅ d s
G G
- IR. What about when we move through our one-loop inductor? There is no electric field in this loop if the resistance of the wire making up the loop is zero. Thus, going around the closed loop clockwise against the current, we have
∫ E^ ⋅^^ d s = −^ ε^ + IR
G G
v (11.4.2)
Now, what is the magnetic flux Φ (^) B through our open surface? First of all, we arrange
the geometry so that the part of the circuit which includes the battery, the switch, and the resistor makes only a small contribution to Φ^ B as compared to the (much larger in area)
part of the open surface which includes our one-loop inductor. Second, we know that is positive in that part of the surface, because current flowing counterclockwise will
produce a magnetic field B pointing out of the page, which is the same direction we have
assumed for. Thus, the dot product
Φ B
G
d A
G
B ⋅ d A > 0
G G
. Note that B
G
is the self magnetic field − that is, the magnetic field produced by the current flowing in the circuit, and not by any external currents.
From Section 11.1, we also see that the magnetic flux Φ (^) B is proportional to I , and may
be written as , where L is the self-inductance which depends on the geometry of
the circuit. The time rate of change of
Φ B = LI
Φ (^) B is just , so that we have from
Faraday's law
L dI dt ( / )
d IR d^ B LdI dt dt
ε
∫ E^^ ⋅^ s = −^ +^ = −^ = −
G G
v (11.4.3)
We can write the governing equation for I ( ) t from above as
dI V IR L dt
∆ = ε− − = (11.4.4)
where the expression has been cast in a form that resembles Kirchhoff's loop rule, namely that the sum of the potential drops around a circuit is zero. To preserve the loop rule, we must specify the “potential drop” across an inductor.
(a)
(b)
Figure 11.4.2 Modified Kirchhoff’s rule for inductors (a) with increasing current, and (b) with decreasing current.
The modified rule for inductors may be obtained as follows: The polarity of the self- induced emf is such as to oppose the change in current, in accord with Lenz’s law. If the rate of change of current is positive, as shown in Figure 11.4.2(a), the self-induced emf ε (^) L sets up an induced current I (^) indmoving in the opposite direction of the current I to
oppose such an increase. The inductor could be replaced by an emf | ε (^) L | = L dI dt | / | = + L dI dt ( / )with the polarity shown in Figure 11.4.2(a). On the other
hand, if dI dt / < 0 , as shown in Figure 11.4.2(b), the induced current I (^) indset up by the
self-induced emf ε (^) L flows in the same direction as I to oppose such a decrease.
We see that whether the rate of change of current in increasing ( ) or decreasing ( ), in both cases, the change in potential when moving from a to b along the direction of the current I is. Thus, we have
dI / dt > 0 dI / dt < 0 Vb − Va = − L d I ( / d t )
Kirchhoff's Loop Rule Modified for Inductors :
If an inductor is traversed in the direction of the current, the “potential change” is − L dI ( / dt ). On the other hand, if the inductor is traversed in the direction opposite of the
current, the “potential change” is + L dI ( / dt ).
Use of this modified Kirchhoff’s rule will give the correct equations for circuit problems that contain inductors. However, keep in mind that it is misleading at best, and at some level wrong in terms of the physics. Again, we emphasize that Kirchhoff's loop rule was
originally based on the fact that the line integral of E
G
around a closed loop was zero.
L
R
τ = (11.4.8)
is the time constant of the RL circuit. The qualitative behavior of the current as a function of time is depicted in Figure 11.4.4.
Figure 11.4.4 Current in the RL circuit as a function of time
Note that after a sufficiently long time, the current reaches its equilibrium value ε / R. The time constant τ is a measure of how fast the equilibrium state is attained; the larger the value of L , the longer it takes to build up the current. A comparison of the behavior of current in a circuit with or without an inductor is shown in Figure 11.4.5 below.
Similarly, the magnitude of the self-induced emf can be obtained as
| | t^ / L
dI L e dt
ε = − = ε −^ τ (11.4.9)
which is at a maximum when t^ = 0 and vanishes as^ t^ approaches infinity. This implies that a sufficiently long time after the switch is closed, self-induction disappears and the inductor simply acts as a conducting wire connecting two parts of the circuit.
Figure 11.4.5 Behavior of current in a circuit with or without an inductor
To see that energy is conserved in the circuit, we multiply Eq. (11.4.7) by I and obtain
I I R^2 LI dI dt
ε = + (11.4.10)
The left-hand side represents the rate at which the battery delivers energy to the circuit. On the other hand, the first term on the right-hand side is the power dissipated in the resistor in the form of heat, and the second term is the rate at which energy is stored in the inductor. While the energy dissipated through the resistor is irrecoverable, the magnetic energy stored in the inductor can be released later.
11.4.3 Decaying Current
Next we consider the RL circuit shown in Figure 11.4.6. Suppose the switch S 1 has been closed for a long time so that the current is at its equilibrium value ε / R. What happens to the current when at t = 0 switches S 1 is opened and S 2 closed?
Applying modified Kirchhoff’s loop rule to the right loop for decreasing current, dI dt / < 0 , yields
| L | 0
dI IR L IR dt
ε − = − − = (11.4.11)
which can be rewritten as
dI dt I L R
Figure 11.4.6 (a) RL circuit with decaying current, and (b) equivalent circuit.
The solution to the above differential equation is
I ( ) t et^ / R
= ε^ −^ τ (11.4.13)
where τ = L / R is the same time constant as in the case of rising current. A plot of the current as a function of time is shown in Figure 11.4.7.
