30. Inductance
Self-Inductance of a Solenoid
Part A
Suppose that the current in the solenoid is
what is the magnetic field
. Within the solenoid, but far from its ends,
due to this current?
Express your answer in terms of
constants (such as ).
, quantities given in the introduction, and relevant
=mu_0*n*I(t)Correct
Note that this field is independent of the radial position (the distance from the axis of
symmetry) as long as it is measured at a point well inside the solenoid.
Part B
What is the magnetic flux
through a single turn of the solenoid?
Express your answer in terms of the magnetic field
introduction, and any needed constants.
, quantities given in the
=B(t)*R^2*piCorrect
Part C
Suppose that the current varies with time, so that
. Find the electromotive
force induced across the entire solenoid due to the change in current through the entire
solenoid.
Express your answer in terms of
, ,
, and
.
=-mu_0*n^2*R^2*pi*Z*dI(t)/dtCorrect
Part D
The self-inductance is related to the self-induced EMF
by the equation
. Find for a long solenoid. (Hint: The self-inductance will always be
a positive quantity.)
Express the self-inductance in terms of the number of turns per length , the physical
dimensions
and
, and relevant constants.
=mu_0*n^2*R^2*pi*ZCorrect
This definition of the inductance is identical to another definition you may have
encountered:
, where
is the magnetic flux due to a current in the inductor.
To see the correspondence you should differentiate both sides of this equation with
respect to time and use Faraday's law, i.e.,
.
Now consider an inductor as a circuit element. Since we are now treating the inductor as
a circuit element, we must discuss the voltage across it, not the EMF inside it. The
important point is that the inductor is assumed to have no resistance. This means that the
net electric field inside it must be zero when it is connected in a circuit. Otherwise, the
current in it will become infinite. This means that the induced electric field deposits
charges on and around the inductor in such a way as to produce a nearly equal and
opposite electric field
such that
! Kirchhoff's loop law defines voltages
only in terms of fields produced by charges (like
), not those produced by changing
magnetic fields (like ). So if we wish to continue to use Kirchhoff's loop law, we must
continue to use this definition consistently. That is, we must define the voltage
alone (note that the integral is from A to B rather than from
B to A, hence the positive sign). Note also that this voltage is no longer the same as the
total difference in electric potential, which is zero, because the net electric field is zero.
So finally, , where we have used
and the definition of .
Part E
Which of the following statements is true about the inductor in the figure in the problem
introduction, where
If
Correct
is the current through the wire?
is positive, the voltage at end A will necessarily be greater than that at end B.
Part F
Now consider the effect that applying an additional voltage to the inductor will have on
the current
already flowing through it (imagine that the voltage is applied to end A,
while end B is grounded). Which one of the following statements is true?
If is positive, then
positive.
Correct
could be positive or negative while
will necessarily be
Note that when you apply Kirchhoff's rules and traverse the inductor in the direction of
current flow, you are interested in
resistor gives a term
, just as traversing a
.
In sum: when an inductor is in a circuit and the current is changing, the changing
magnetic field in the inductor produces an electric field. This field opposes the change in
current, but at the same time deposits charge, producing yet another electric field. The net
effect of these electric fields is that the current changes, but not abruptly. The "direction
of the EMF" refers to the direction of the first, induced, electric field.
Energy Stored in an Inductor
Part A
What inductance would be needed to store energy
carrying current
(kilowatt-hours) in a coil
?
=240 HCorrect
Growth of Current in an L-R Circuit
Part A
What is the differential equation governing the growth of current in the circuit as a
function of time after
?
Express the right-hand side of the differential equation for
and .
=V_b/L-R*I(t)/LCorrect
Part B
in terms of
,
,
,
We know that the current in the circuit is growing. It will approach a steady state after a
long time (as tends to infinity), which implies that the differential term in our circuit
equation will tend to zero, hence simplifying our equation. The current around the circuit
will tend to an asymptotic value
. What is
Express your answer in terms of
introduction.
and
?
and other constants and variables given in the
=V_b/RCorrect
Part C
Solve the differential equation obtained in Part A for the current
after the switch is closed at
Express your answer in terms of
as a function of time
.
,
, and . Use the notation exp(x) for
.
=V_b/R*(1-exp(-R*t/L))Correct
The current approaches its asymptotic/steady-state value exponentially, i.e., usually very
fast. However, the time constant depends on the actual values of the inductance and
resistance.
The R-L Circuit: Responding to Changes
Part A
Immediately after the switch is closed, what is the current in the circuit?
zeroCorrect
Part B
Immediately after the switch is closed, what is the voltage across the resistor?
zeroCorrect
Part C
Immediately after the switch is closed, what is the voltage across the inductor?
Correct
Part D
Shortly after the switch is closed, what is the direction of the current in the circuit?
counterclockwise
Correct
Part E
Shortly after the switch is closed, what is the direction of the induced EMF in the
inductor?
clockwise
Correct
The induced EMF does not oppose the current; it opposes the change in current. In this
case, the current is directed counterclockwise and is increasing; in other words, the time
derivative of the counterclockwise current is positive, which is why the induced EMF
is directed clockwise.
Part F
Eventually, the process approaches a steady state. What is the current in the circuit in the
steady state?
Correct
Part G
What is the voltage
zero
Correct
Part H
What is the voltage
Correct
across the inductor in the steady state?
across the resistor in the steady state?
Part I
Now that we have a feel for the state of the circuit in its steady state, let us obtain the
expression for the current in the circuit as a function of time. Note that we can use the
loop rule (going around counterclockwise):
.
Note as well that
and
. Using these equations, we can get, after some
rearranging of the variables and making the subsitution
,
.
Integrating both sides of this equation yields
.
Use this last expression to obtain an expression for
that
. Remember that
and
.
Express your answer in terms of ,
, and . You may or may not need all these
variables. Use the notation exp(x) for
.
=EMF/R*(1-exp(-R*t/L))Correct
Part J
Just as in the case of R-C circuits, the steady state here is never actually reached: The
exponential functions approach their limits asymptotically as
. However, it usually
does not take very long for the value of to get very close to its presumed limiting value.
The next several questions illustrate this point.
Note that the quantity
has dimensions of time and is called the time constant (you
may recall similar terminology applied to R-C circuits). The time constant is often
denoted by . Using , one can write the expression
as
.
Find the ratio of the current
at time
to the maximum current
Express your answer numerically, using three significant figures.
.
=0.9980Correct
Part K
Find the time it takes the current to reach 99.999% of its maximum value.
Express your answer numerically, in units of . Use three significant figures.
=11.50 Correct
Part L
Find the time it takes the current to reach 99.999% of its maximum value. Assume that
ohms and
millihenrys.
Express your answer in seconds, using three significant figures.
=5.760×10−2 secondsCorrect
It does not take long at all! However, the situation can be different if the inductance is
large and the resistance is small. The next example illustrates this point.
Part M
Find the time it takes the current to reach 99.999% of its maximum value. Assume that
ohms and
henrys.
Express your answer in seconds, using three significant figures.
=5760 secondsCorrect
This is more than an hour and a half! Nobody would wait that long. Let us see what
change would occur in such a circuit over a shorter period of time.
Part N
What fraction of the maximum value will be reached by the current one minute after the
switch is closed? Again, assume that
Use three significant figures in your answer.
ohms and
henrys.
=0.1130Correct
This is only 11.3% of the maximum value of the current!
Now consider a different situation. After switch
has been closed for a long time, it is
opened; simultaneously, switch is closed, as shown in the figure. This effectively
removes the battery from the circuit.
The questions below refer to the time immediately after switch
is closed.
is opened and switch
Part O
What is the direction of the current in the circuit?
counterclockwise
Correct
Part P
What is happening to the magnitude of the current?
The current is decreasing.
Correct
Part Q
What is the direction of the EMF in the inductor?
counterclockwise
Correct
Part R
Which end of the inductor has higher voltage (i.e., to which end of the inductor should
the positive terminal of a voltmeter be connected in order to yield a positive reading)?
left
Correct
Part S
For this circuit, Kirchhoff's loop rule gives
.
Note that
, since the current is decreasing. Use this equation to obtain an
expression for
.
Express your answer in terms of , , and
. Use exp(x) for
.
=EMF/R*exp(-R*t/L)Correct
Oscillations in an LC circuit.
Initially, the switch is open, and the capacitor has a charge . The switch is then closed,
and the changes in the system are observed. It turns out that the equation describing the
subsequent changes in charge, current, and voltage is very similar to that of simple
harmonic motion, studied in mechanics. To obtain this equation, we will use the law of
conservation of energy.
Initially, the entire energy of the system is stored in the capacitor. When the circuit is
closed, the capacitor begins to discharge through the inductor. As the charge of the
capacitor decreases, so does its energy. On the other hand, as the current through the
inductor increases, so does the energy stored in the inductor. Assuming no heat loss and
no emission of electromagnetic waves, energy is conserved, and at any point in time, the
sum of the energy stored in the capacitor
constant
and the energy stored in the inductor
:
,
is a
where is the charge on the capacitor and is the current through the inductor ( and are
functions of time, of course). For this problem, take clockwise current to be positive.
Part A
Using the expression for the total energy of this system, it is possible to show that after
the switch is closed,
,
where is a constant. Find the value of the constant .
Express your answer in terms of and
.
=1/(L*C)Correct
This expression can also be obtained by substituting
into the equation obtained
from Kirchhoff's loop rule. It is always good to solve a problem in more than one way!
Part B
From mechanics, you may recall that when the acceleration of an object is proportional to
its coordinate,
,
such motion is called simple harmonic motion, and the coordinate depends on time as
, where , the argument of the harmonic function at
, is called
the phase constant.
Find a similar expression for the charge
on the capacitor in this circuit. Do not forget
to determine the correct value of based on the initial conditions described in the
problem.
Express your answer in terms of
, , and
. Use the cosine function in your answer.
=q_0*cos(t/sqrt(L*C))Correct
Part C
What is the current
in the circuit at time after the switch is closed?
Express your answer in terms of
, ,
, and other variables given in the introduction.
=q_0/sqrt(L*C)*sin(t/sqrt(L*C))Correct
Part D
Recall that the top plate of the capacitor is positively charged at
. In what direction
does the current in the circuit begin to flow immediately after the switch is closed?
clockwise
Correct
Part E
Immediately after the switch is closed, what is the direction of the EMF in the inductor?
(Recall that the direction of the EMF refers to the direction of the back-current or the
induced electric field in the inductor.)
counterclockwiseCorrect
In the remaining parts, assume that the period of oscillations is 8.0 milliseconds. Also,
keep in mind that the top plate of the capacitor is positively charged at
.
Part F
At what time does the current reach its maximum value for the first time?
2.0 msCorrect
Part G
At what moment does the EMF become zero for the first time?
2.0 Correct
Part H
At what moment does the current reverse direction for the first time?
4.0 msCorrect
Part I
At what moment does the EMF reverse direction for the first time?
2.0 msCorrect
Part J
At what moment does the energy stored in the inductor reach its maximum for the first
time?
2.0 msCorrect
Part K
At what time do the capacitor and the inductor possess the same amount of energy for
the first time?
Express your answer in milliseconds.
=1.00 Correct
Part L
What is the direction of the current in the circuit 22.0 milliseconds after the switch is
closed?
counterclockwiseCorrect
Part M
What is the direction of the EMF in the circuit 42.0 milliseconds after the switch is
closed?
The EMF is zero
Correct