Week 11 - Inductance
November 5, 2013
Exercise 11.1: Discussion Questions
a) A transformer consists basically of two coils in close proximity but not in electrical contact. A current
in one coil magnetically induces an emf in the second coil, with properties that can be controlled by
adjusting the geometry of the two coils. Such a device will work only with alternating current,
however, and not with direct current. Explain why.
Solution:
A transformer only works with alternating current because it is dependent upon the phenomenon of
mutual inductance. Here the emf in the second coil due to the first is given by
ε2 = −M
di1
,
dt
(1)
and thus, for there to be an emf, there has to be a changing current in the first coil. The magnitude
of the emf for a given current change is determined by the mutual inductance M , which depends only
on the geometry of the coils.
b) Suppose there is a steady current in an inductor. If you attempt to reduce the current to zero
instantaneously by quickly opening a switch, an arc1 can appear at the switch contacts. Why? Is it
physically possible to stop the current instantaneously? Explain.
Solution:
The arc is formed because of the self inductance in the circuit. When you open the switch the rate
of change of the current become extremely high and therefore the emf generated becomes enormous.
This emf forces the current across the gap between the switch terminals. Mathematically this is seen
with the equation
ε = −L
di
dt
(2)
where di/dt becomes very large and negative as you open the switch, and consequently ε becomes
huge and forces the current across the gap. Since all circuits have some self conductance it is not
possible to stop any current instantaneously.
1 An
electric arc is a luminous discharge of current that is formed when a current jumps a gap in a circuit between two
electrodes.
1
Exercise 11.2: Inductance of a Solenoid
A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the
inductance of this solenoid is given by the equation L = µ0 AN 2 /l. Assume that the magnetic field is
uniform inside the solenoid and zero outside. (Your answer is approximate because B is actually smaller
at the ends than at the center. For this reason your answer is actually an upper limit on the inductance.)
Solution:
For the case of an ’infinitely long’ solenoid the magnetic field is is B = µ0 nI where n is the number of
turns per unit length. Now we use this field to approximate the inductance of a finite solenoid. A finite
solenoid has a finite length l and finite number of turns N so we can write
B = µ0 nI =
µ0 N I
.
l
(3)
For self inductance the flux trough the area enclosed by the turns of wire in the solenoid is
φB = B(N A) =
µ0 AN 2
I
l
(4)
and here we see the claim that the flux φB is always proportional to the current I. The proportionality
constant is the self inductance L
L=
µ0 AN 2
.
l
(5)
Exercise 11.3: A RL-Circuit
In the circuit shown in figure 1, E = 50.0 V, R1 = 40.0 Ω,R2 = 30.0 Ω and L = 0.70 H. Switch S is closed
at t = 0. Just after the switch is closed:
a) What is the potential difference vab across the resistor R1 ?
Solution: Just after the switch is closed the current trough the inductor is zero because it hasn’t had
any time to increase (Recall how the current grows in a simple RL-circuit). Therefore by applying
the loop rule to the upper loop
E − vab = 0 ⇒ vab = E.
(6)
vab = E.
(7)
Answer:
2
Figure 1
b) Which point, a or b, is at a higher potential.
Solution:
As the current travels trough R1 the voltage drops by IR1 , therefore b is at a lower potential than a.
Answer: b is at the lower potential.
c) What is the potential difference vcd across the inductor L?
Solution:
There is no current trough R2 so by the loop rule
E − i2 R2 − vcd = E − vcd = 0 ⇒ vcd = E.
(8)
vcd = E.
(9)
Answer:
d) Which point, c or d, is at a higher potential?
Solution:
The voltage increases when we go from b to a trough the emf, so it must therefore drop as we go
trough the inductor from c to d. Thus c is at the higher potential.
3
Answer: c is at the higher potential.
The switch is then left closed a long time and then opened. Just after the switch is opened... Repeat (a)
to (d).
Solution:
(a) Since the switch has been open for a long time, i2 is not changing anymore (di2 /dt = 0), so vcd = 0.
Therefore by the loop rule
E = vR2 = i2 R2 ⇒ i2 =
E
.
R2
(10)
(b) Now when the switch suddenly opens the current trough the inductor has not had time to change so
the current trough the lower part of the circuit (with S open) is i2 . This current travels from b to a, so
b is therefore at a higher potential.
Answer:
i2 =
E
R2
(11)
a is at the higher potential.
Exercise 11.4: An Electromagnetic Car Alarm
Your Lot’s invention is a car alarm that produces sound at a particularly annoying frequency of 3500 Hz.
To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency.
That’s why your design includes an inductor and a capacitor in series. The maximum voltage across the
capacitor is to be 12.0 V (the same voltage as that of the car battery). To produce a sufficiently loud
sound, the capacitor must store 0.0160 J of energy. What values of capacitance and inductance should
you chose for your car-alarm circuit?
Solution:
We seek the relation between a frequency of a LC-circuit and it’s capacitance and inductance
p
ω = 1/LC. Unless you are very adept at remembering formulas we recommend focusing on learning
this method. The voltage around the loop has to be zero, for a capacitor and an inductor of respective
voltages q/C and Ldi/dt this means that
L
di
q
+
= 0.
dt C
(12)
Now the charge on the capacitor is related to the current by I = dq/dt, therefore in terms of q we can
write
1
d2 q
+
q = 0.
2
dt
LC
A good guess for this differential equation is q(t) = Q cos (ωt + φ). Substitution yields
4
(13)
1
2
cos (ωt + φ) = 0,
Q −ω cos (ωt + φ) +
LC
and we see that our guess is only a solution if ω =
the textbook).
1
LC
(14)
which is our desired result (you can look it up in
Now for our car-alarm frequency; ω is the angular frequency and this is related to the actual frequency
ω = 2πf . Therefore we get
f=
ω
1
√
.
=
2π
2π LC
(15)
We have to find the appropriate values of C and L. We know the maximum voltage across the capacitor as
well as it’s energy. Therefore we can use the relation for the energy of a capacitor UE = Q2 /2C = V 2 C/2
to find it’s capacitance. We get
C=
2 × 0.0160
2UE
=
F = 222 µF
V2
(12.0)2
(16)
.
Now we can determine the necessary inductance by our equation for the frequency. Rearranging we get
L=
1
2πf
2
1
=
C
1
2π × 3500
2
1
H = 9.3 µH.
222 × 10−6
(17)
Answer:
L=
1
2π × 3500
2
1
H = 9.3 µH.
222 × 10−6
(18)
Exercise 11.5: Solar Magnetic Energy
Magnetic fields within a sunspot can be as strong as 0.4 T. (By comparison, the earth’s magnetic field
is about 1/10000 as strong.) Sunspots can be as large as 25000 km in radius. The material in a sunspot
has a density of about 3 × 10−4 kg/m3 . Assume µ for the sunspot material to be µ0 . If 100% of the
magnetic-field energy stored in a sunspot could be used to eject the sunspot’s material away from the
sun’s surface, at what speed would that material be ejected? Compare to the sun’s escape speed, which
is about 6 × 105 m/s. Hint: Calculate the kinetic energy the magnetic field could supply to 1m3 of sunspot
material.
Solution:
Under the assumption that the permeability is µ0 the magnetic field energy density is given by
uB =
B2
2µ0
5
(19)
and therefore energy in 1 m3 is
U=
B2
× 1 m3 .
2µ0
(20)
Now if this energy is 100% converted to kinetic energy of the sunspot material, we can set up a conservation of energy equation for 1 m3 of this material. When all the energy is converted to kinetic energy
we have
1
B2
× 1 m3 = mv 2
2µ0
2
(21)
or equivalently
v=√
B
.
µ0 m
(22)
1 m3 of sunspot material have a mass m = 3 × 10−4 kg/m3 × 1 m3 . We then get a velocity
v=√
0.4
B
=√
m/s = 20601.6 m/s.
µ0 m
4π × 10−7 × 3 × 10−4
(23)
This is only 3.4% of the escape velocity of the sun, so the magnetic field energy alone is not enough to
cause ’coronal mass ejections’ which is a massive burst of particles being ejected into space. It remains
a mystery exactly how the particles get that much energy.2
Answer:
v == 20601.6 m/s.
(24)
This is only 3.4% of the escape velocity of the sun, so the magnetic field energy alone is not enough
cause ’coronal mass ejections’ which is a massive burst of particles being ejected into space. It remains
a mystery exactly how the particles get that much energy.3
Exercise 11.6: Inductance of a Coaxial Cable
A small solid conductor of radius a is supported by insulating, nonmagnetic disks on the axis of a thinwalled tube with inner radius b. The inner and outer conductors carry equal currents i in opposite
directions.
a) Use Ampere’s law to find the magnetic field at any point in the volume between the two conductors.
Solution:
2 Solar
3 Solar
flare Wikipedia.
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6
Figure 2
We construct a circular integration path with radius r centered at the axis of the cable. Then by
Ampere’s law
I
B · dl = B2πr = µ0 i
(25)
or equivalently
B=
µ0 i
.
2πr
(26)
B=
µ0 i
.
2πr
(27)
Answer:
b) Write the expression for the flux dφB through a narrow strip of length l parallel to the axis, of width
dr, at a distance r from the axis of the cable and lying in a plane containing the axis.
Solution:
The narrow strip is shown in figure 3. The flux trough this strip will be the magnetic field (which is
perpendicular to the strip at every point) times the area of the strip. I.e. it will be
dφB = B (ldr) = l
µ0 i
dr.
2πr
(28)
dφB = B (ldr) = l
µ0 i
dr.
2πr
(29)
Answer:
7
Figure 3
c) Integrate your expression from part (b) over the volume between the two conductors to find the total
flux produced by the current i in the central conductor.
Solution:
To calculate the entire flux we have to integrate this expression from a to b.
Z
φB =
b
dφ = l
a
µ0 i
2π
b
Z
a
dr
µ0 i
=l
ln
r
2π
b
.
a
(30)
Answer:
b
Z
φB =
dφ = l
a
µ0 i
ln
2π
b
.
a
(31)
d) Show that the inductance of a length l of the cable is
L=l
µ0
ln
2π
b
.
a
(32)
Solution:
The inductance is the constant of proportionality between φB and i. Looking at our expression for
the flux
φB = l
µ0
ln
2π
b
i = Li
a
we see that
8
(33)
L=l
µ0
ln
2π
b
.
a
(34)
e) Use equation 32 to calculate the energy stored in the magnetic field for a length l of the cable.
Solution:
We know that the energy stored in the magnetic field of an inductor is 12 Li2 . Therefore we get
U =l
µ0 i2
ln
4π
b
.
a
(35)
U =l
µ0 i2
ln
4π
b
.
a
(36)
Answer:
f) Verify that this is the correct expression for the energy by also calculating the total energy via the
B2
energy density formula uB = 2µ
. Hint: Calculate the energy of a thin, cylindrical, shell of length l
0
and radius r placed in between the two conductors.
Solution:
We found the magnetic field in between the two conductors in (a). The result was
B=
µ0 i
.
2πr
(37)
The corresponding energy density is therefore given by
B2
uB =
=
2µ0
µ0 i 2
2πr
2µ0
=
µ0 i2
.
8π 2 r2
(38)
Now to find the energy of a thin, cylindrical shell of length l and radius r we must multiply the energy
density with the volume of the cylindrical shell. The volume of such a shell is given by dV = l2πrdr
and therefore
dU = uB dV =
µ0 i2
µ0 i2
2πrdr
=
l
dr.
8π 2 r2
4πr
(39)
such that the total energy contained in the magnetic field within the conductors is
µ0 i2
l
4π
Z
a
b
dr
µ0 i2
b
=l
ln
r
4π
a
which indeed agrees with the expression from (e).
9
(40)