Physics 4
Inductance
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Inductance
Mutual Inductance of two coils:
Some of the magnetic flux through one
coil also passes through the other coil,
inducing a voltage.
Inductance is magnetic flux/current.
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Self-Inductance
Changing current through the wires in a
coil will induce a voltage that opposes
the CHANGE in the current.
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Magnetic Field Energy
When an inductor has a steady current, it stores potential energy.
This leads to a general formula for potential
energy stored in any magnetic field:
𝐵2
𝑢=
2𝜇0
This formula is for magnetic energy density,
Which is energy per unit volume.
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Example: A solenoid 25.0cm long and with a cross-sectional area of 0.500cm2
contains 400 turns of wire and carries a current of 80.0A.
Calculate:
(a) the magnetic field in the solenoid.
(b) the energy density in the magnetic field if the solenoid is air-filled.
(c) the total energy contained in the coil’s magnetic field (assume uniform field).
(d) The inductance of the solenoid.
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Example: A solenoid 25.0cm long and with a cross-sectional area of 0.500cm2
contains 400 turns of wire and carries a current of 80.0A.
Calculate:
(a) the magnetic field in the solenoid.
(b) the energy density in the magnetic field if the solenoid is air-filled.
(c) the total energy contained in the coil’s magnetic field (assume uniform field).
(d) The inductance of the solenoid.
Solution:
𝐵𝑠𝑜𝑙𝑒𝑛𝑜𝑖𝑑 = 𝜇0 𝑛𝐼 = 4𝜋 ∙ 10−7
400 𝑡𝑢𝑟𝑛𝑠
0.25𝑚
80𝐴 = 0.16𝑇
𝐵2
(0.16𝑇) 2
𝑢=
=
= 10,294𝑚𝐽3
2𝜇0 2 ∙ 4𝜋 ∙ 10−7
𝑈𝑚𝑎𝑔 = 𝑢 ∙ 𝑣𝑜𝑙𝑢𝑚𝑒 = 10,294𝑚𝐽3 0.25𝑚 0.5𝑐𝑚2
1𝑚
100𝑐𝑚
2
= 0.129𝐽
For part d) we can use the formula for energy in an inductor:
𝑈 = 12𝐿𝐼 2 → 𝐿 =
2𝑈 2(0.129𝐽)
=
= 4 ∙ 10−5 𝐻 = 400𝜇𝐻
2
2
𝐼
(80𝐴)
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R-L Circuit
When connected in a circuit with a
resistor, an inductor will have the
effect of slowing down changes in the
current through the resistor.
When the current is steady (the switch
has been closed for a long time), the
inductor has no effect, but there is
potential energy stored in the inductor.
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R-L Circuit
If switch S1 is closed in the circuit,
current will begin to flow through the
resistor and inductor as shown. This
increasing current will induce current
to flow the opposite direction, slowing
the growth of the current.
We can write down a formula for the
current as a function of time:
𝑖 𝑡 =
𝜀
1 − 𝑒−
𝑅
𝑅
𝐿 𝑡
𝐿
The quantity 𝜏 = 𝑅 is called the “time
constant” for this exponential decay.
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R-L Circuit
Once the current reaches a steady
value we can flip the switches,
opening S1 and closing S2. Then
current will keep flowing for while as
the inductor opposes this decreasing
current.
A similar formula describes this
decaying current as a function of time:
𝑖 𝑡 = 𝐼0 𝑒 −
𝑅
𝐿 𝑡
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Example: A 35.0V battery with negligible internal resistance, a 50.0Ω resistor and a 1.25mH
inductor are connected in series with an open switch. The switch is suddenly closed.
(a) How long after closing the switch will the current through the inductor reach half of its
maximum value?
(b) How long after closing the switch will the energy stored in the inductor reach half its
maximum value?
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Example: A 35.0V battery with negligible internal resistance, a 50.0Ω resistor and a 1.25mH
inductor are connected in series with an open switch. The switch is suddenly closed.
(a) How long after closing the switch will the current through the inductor reach half of its
maximum value?
(b) How long after closing the switch will the energy stored in the inductor reach half its
maximum value?
As soon as the switch is closed, current begins to flow around the circuit, increasing toward
a maximum value given by Ohm’s Law. Here is the formula:
𝑖 𝑡 =
𝜀
1 − 𝑒−
𝑅
𝑅
𝐿 𝑡
We want to find the time when the current is half of the maximum.
𝑖 𝑡 =
1
2
𝜀
𝜀
=
1 − 𝑒−
𝑅
𝑅
− 𝑅 𝐿 𝑡 = ln
1
2
𝑅
𝐿
→𝑡=−
∙ ln
𝑅
𝐿 𝑡
1
2
1
2
→ = 1−
𝑅
𝑒− 𝐿 𝑡
1.25 ∙ 10−3 𝐻
=−
∙ ln
50Ω
1
2
→
𝑅
𝑒− 𝐿 𝑡
1
=
2
= 1.73 ∙ 10−5 𝑠 = 17.3𝜇𝑠
For part b) we want the energy to be half of its maximum, so use the energy formula:
𝑈 = 12𝐿𝑖 2 = 12 12𝐿𝐼 2 → 𝑖(𝑡) =
𝐼
2
Using the formula for current again:
1
2
= 1 − 𝑒−
𝑅
𝐿 𝑡
→ 𝑡 = 30.7𝜇𝑠
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L-C Circuit
A circuit containing a capacitor and an inductor will exhibit an oscillating
current, with potential energy transferring back and forth.
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L-C Circuit
The oscillations in an L-C circuit should
look familiar. This situation is directly
analogous to an undamped mass-spring
system that we saw previously.
All of the formulas we developed for that
case are repeated here, with charge, q,
taking the place of displacement, x. The
capacitor is related to the spring constant,
and the inductance is like mass.
To add in the damping, we just include a
resistor in the circuit…
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
(a) What is the maximum charge on the capacitor?
We can use energy for this if we want to. When all the energy is in the inductor it will have
maximum current. When all the energy is in the capacitor it will have maximum charge.
𝑈𝑚𝑎𝑔 = 12𝐿𝐼 2 = 12(85 ∙ 10−3 𝐻)(0.85 ∙ 10−3 )2 = 0.307𝑛𝐽
𝑈𝑒𝑙𝑒𝑐 =
2
1𝑄
2 𝐶
= 0.307𝑛𝐽 → 𝑄 = 44.3𝑛𝐶
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
b) We can use energy again, or we can use the formula for the charge as a function of time.
𝐸𝑡𝑜𝑡𝑎𝑙 =
1 2
𝐿𝑖
2
+
2
1𝑞
2 𝐶
Total energy can be found from max current or max charge. Should be the same either way.
We can solve for the charge when the current is as given:
𝑞2
1
1
−8
−3
−3
2
𝐸𝑡𝑜𝑡𝑎𝑙 = 3.07 ∙ 10 𝐽 = 2 85 ∙ 10 𝐻 0.5 ∙ 10 𝐴 + 2
→ 𝑞 = 3.58 ∙ 10−7 𝐶 = 358𝑛𝐶
3.2 ∙ 10−6 𝐶
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Example: In an L-C circuit, L=85.0mH and C=3.20μF. During the oscillations the maximum
current in the inductor is 0.850mA.
(a) What is the maximum charge on the capacitor?
(b) What is the magnitude of the charge on the capacitor at an instant when the current in
the inductor has magnitude 0.500mA?
(c) How long does it take for the capacitor to go from maximum charge to zero charge?
To go from no charge to fully charged is a quarter of a cycle, so we need to find the period of
the oscillation. We have a formula for angular frequency:
𝜔=
1
= 1917𝑟𝑎𝑑
𝑠
𝐿𝐶
Rearrange this to get the period, then divide by 4:
𝑇=
2𝜋
= 0.00328𝑠 → 14𝑇 = 0.82𝑚𝑠
𝜔
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L-R-C Series Circuit
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L-R-C Series Circuit
Formulas for this case are developed in the same way as the L-C
circuit, we just include an extra term involving resistance:
𝑑2 𝑞 𝑅 𝑑𝑞
1
+
+
𝑞=0
2
𝑑𝑡
𝐿 𝑑𝑡 𝐿𝐶
Solving this differential equation gives a general solution:
𝑞=
− 𝑅 2𝐿 𝑡
𝐴𝑒
𝑐𝑜𝑠
1
𝑅2
− 2𝑡+𝜑
𝐿𝐶 4𝐿
This solution is for the underdamped case: R2<4L/C
The angular frequency in this case is:
𝜔′ =
1
𝑅2
− 2
𝐿𝐶 4𝐿
Notice this is less than the frequency in
the undamped L-C circuit – the resistor
slows down the oscillations.
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Example: Assume the switch has been in the position shown in the figure for a long time (so
the capacitor is fully charged and no current is flowing). When the switch is moved (to
connect points a and d in the figure), find the following:
a) The initial charge on the capacitor, and initial total energy in this system.
b) The frequency of oscillation for this circuit.
c) The maximum current through the inductor, and the time when that current is first
achieved.
Assume the following values: ε=10.0V, R=1kΩ, C=1μF, L=2H.
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Example: Assume the switch has been in the position shown in the figure for a long time (so
the capacitor is fully charged and no current is flowing). When the switch is moved (to
connect points a and d in the figure), find the following:
a) The initial charge on the capacitor, and initial total energy in this system.
b) The frequency of oscillation for this circuit.
c) The maximum current through the inductor, and the time when that current is first
achieved.
Assume the following values: ε=10.0V, R=1kΩ, C=1μF, L=2H.
a) At the beginning, the capacitor is fully charged, so the voltage matches the battery.
𝑄 = 𝐶 ∙ 𝑉 = 1𝜇𝐹 10𝑉 = 10𝜇𝐶
𝑈𝑒𝑙𝑒𝑐 = 12𝐶𝑉 2 = 12 1𝜇𝐹 10𝑉
2
= 50𝜇𝐽
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Example: Assume the switch has been in the position shown in the figure for a long time (so
the capacitor is fully charged and no current is flowing). When the switch is moved (to
connect points a and d in the figure), find the following:
a) The initial charge on the capacitor, and initial total energy in this system.
b) The frequency of oscillation for this circuit.
c) The maximum current through the inductor, and the time when that current is first
achieved.
Assume the following values: ε=10.0V, R=1kΩ, C=1μF, L=2H.
a) At the beginning, the capacitor is fully charged, so the voltage matches the battery.
𝑄 = 𝐶 ∙ 𝑉 = 1𝜇𝐹 10𝑉 = 10𝜇𝐶
𝑈𝑒𝑙𝑒𝑐 = 12𝐶𝑉 2 = 12 1𝜇𝐹 10𝑉
2
= 50𝜇𝐽
b) Frequency for an underdamped system:
′
𝜔 =
1
𝑅2
−
=
𝐿𝐶 4𝐿2
1
(1000Ω)2
−
= 661𝑟𝑎𝑑
−6
2
𝑠
(2𝐻)(10 𝐹)
4(2𝐻)
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Example: Assume the switch has been in the position shown in the figure for a long time (so
the capacitor is fully charged and no current is flowing). When the switch is moved (to
connect points a and d in the figure), find the following:
a) The initial charge on the capacitor, and initial total energy in this system.
b) The frequency of oscillation for this circuit.
c) The maximum current through the inductor, and the time when that current is first
achieved.
Assume the following values: ε=10.0V, R=1kΩ, C=1μF, L=2H.
c) Maximum current in inductor happens when the capacitor discharges – this is ¼ cycle.
2𝜋
→ 𝑇 = 0.0095𝑠 = 9.5𝑚𝑠
𝑇
at t= ¼ (9.5ms)=2.4ms
𝜔′ = 661𝑟𝑎𝑑
=
𝑠
Imax
We will need to put this time into the formula for current, which
is the derivative of the formula for charge on the capacitor.
𝑞 = 𝑄𝑒 −
𝑅
2𝐿 𝑡 𝑐𝑜𝑠
𝑖 = − 𝑅 2𝐿 𝑄𝑒 −
𝑅
𝜔′𝑡
2𝐿 𝑡 𝑐𝑜𝑠
𝑅
2𝐿 𝑡 𝑠𝑖𝑛
− 1000Ω 2 2𝐻
2.4𝑚𝑠
𝜔′ 𝑡 − 𝜔′ 𝑄𝑒 −
𝑖 2.4𝑚𝑠 = − 661𝑟𝑎𝑑
10𝜇𝐶 𝑒
𝑠
𝜔′ 𝑡
= 3.63𝑚𝐴
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