Summary
Lecture 21
Electromagnetic induction can induced electric current in conductors called
eddy currents; these currents will induce a magnetic field which can be then
detected.
A transformer consists of a primary (input) and secondary (output) coils and
Is used to transform low or high AC voltage to high or low, respectively.
Inductance
LR Circuits
Used phenomenon: changing magnetic flux
induces voltage in the secondary coil.
ε = −N
ΔΦB
Δt
VS N S
=
VP N P
Ratio of EMF (voltages) equals to ratio
of the number of turns in each coil:
In a step-up transformer VS > VP and NS > NP.
In a step-down transformer VS < VP and NS < NP.
More equations for
transformers:
Physics 112, Spring 2010, Mar 3, Lecture 21
PP = PS
VP I P = VS I S
VS I P
=
VP I S
2
Physics 112, Spring 2010, Mar 3, Lecture 21
Transformer
Resistors, Capacitors, and Inductors
A transformer is a device that:
Resistors: resistance (ohm)
1) operates on either DC or AC
Capacitors: capacitance (farad)
2) operates only on AC
3) operates only on DC
4) depends on its parameters
Inductor: inductance (henry)
R
C
L
Transformer
In a transformer, the power input
Resistors, capacitors, and inductors
are widely used in electronic circuits.
1) is larger than the power output
2) is equal to the power output
3) is smaller than the power output
Any coil can serve as an inductor.
4) can be either larger or smaller than the power output
Physics 112, Spring 2010, Mar 3, Lecture 21
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Mutual Inductance
Self-Inductance
If the current flowing in a single coil changes, an EMF (called back EMF)
is created in the coil.
Mutual Inductance (e.g. a transformer):
If two coils are placed near each other and a
changing current flows through one coil, it
induces an EMF in the other.
Back EMF tends to block
the current as it enters
the inductor at A
+
-
-
+
Mutual inductance, M.
ΔI
ε 2 = −M 1
Δt
Back EMF tries to
increase the current.
An inductor resists any change in the current.
and vice versa:
ΔI
ε1 = −M 2
Δt
FIGURE 21-31.
Induced EMF in inductor:
ε = −L
The units of mutual inductance: V·s/A = Ω·s = henry (H) [after Joseph Henry].
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Physics 112, Spring 2010, Mar 3, Lecture 21
Units: henry, H.
1 H = 1 V·s/A = 1 Ω·s.
ΔI
Δt
L ⇒ self inductance.
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Physics 112, Spring 2010, Mar 3, Lecture 21
Solenoid Inductance
Inductance
Determine a formula for the self-inductance L of solenoid (long coil) and
calculate the value of L if
What happened with the inductance of a coil if we will place a rod of
iron inside of the coil?
N =100, l = 5 cm, A =0.3 cm3 .
ε = −N
ΔΦ B
L=N
ΔI
ΦB =
A
ΔΦ B
ΔI
= −L
Δt
Δt
ΔΦ B =
l
L=
μ0 N A
2
l
=
B=
μ0 NΔIA
l
μ 0 NI
2) inductance decrease
4) need to know the initial inductance
l
Solenoid
inductance:
L=
L=
μ0 N 2 A
2
Physics 112, Spring 2010, Mar 3, Lecture 21
−5
μ0 N 2 A
L=
l
μN 2 A
L=
l
μiron N 2 A
l
l
The constant μ0 is called the “permeability
of free space” or “magnetic constant”.
(4π × 10 T ⋅ m / A)(100) (3 ×10 m )
= 7.5 μH
5 × 10 − 2 m
−7
1) inductance will be the same
3) inductance increase
l
Φ B = BA
μ0 NIA
N loops
2
μ0 = 4π ×10 −7 T ⋅ m / A
7
The constant μ is called “magnetic
permeability”, depends on a material.
μiron = 6 × 10 −3 T ⋅ m / A
Physics 112, Spring 2010, Mar 3, Lecture 21
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Energy Stored in Inductor
1
energy = U E = CV 2
2
Energy stored in a capacitor:
(due to electric field)
Energy stored in an inductor:
energy = U B =
(due to magnetic field)
B = μ 0 NI / l
L = μ0 N A / l
2
RL Circuit Demo
DC source is connected to
a inductor and resistor (the
bulb) is connected in series
(switch at 1).
I
1 2
LI
2
2
I
I = Bl / μ 0 N
1
2
1
1 ⎛ μ N A ⎞⎛ Bl ⎞
1 B2
⎟⎟ =
⎟⎟⎜⎜
energy = U B = LI 2 = ⎜⎜ 0
Al
2
2 ⎝ l ⎠⎝ μ 0 N ⎠
2 μ0
2
t
If the battery is removed from
the circuit (switch at 2):
1 B2
energy density = u B =
2 μ0
Energy density:
I
I
2
Energy stored in an inductor if
ferromagnetic material is present inside:
energy density = u B =
2
1B
2 μ
1
μ 0 is the permeability of free space,
μ is the magnetic permeability of a material.
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Physics 112, Spring 2010, Mar 3, Lecture 21
t
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Physics 112, Spring 2010, Mar 3, Lecture 21
LR Circuit: Connected to Battery
Solenoid Time Constant
DC source is connected to a inductor and resistor connected in series.
In LR circuit, an inductor has an inductance of 600 mH and a resistor has a
resistance of 0.5 Ω.
Switch at 1: current, I, changes rapidly at first.
1. The time constant for this circuit:
Large back EMF develops across L and opposes the increasing current.
Initially most of the voltage drop is across the inductor.
τ =
With time, I increases less rapidly; eventually all the voltage drop is across R.
L
600 × 10 −3 H
=
= 1 .2 s
R
0 .5 Ω
2. How long it would take for the current to go from zero to 80%:
(
I = I max 1 − e − t / τ
2
When t = τ :
1 − 0 .8 = e − t / τ
(
)
) = 0.63
I = I max 1 − e − t / τ
(1 − e
(
0 .8 I max = I max 1 − e − t / τ
)
0 .8 = 1 − e − t / τ
-
+
1
Current in circuit
at time t:
)
− t /τ
I max =
V
R
τ =
L
R
0 .2 = e − t / τ
ln 0 .2 = ln e − t / τ
− 1 .61 = −
t
τ
t = (1.61)(τ ) = (1.61) = (1.61)(1.2 s) = 1.93 s
I = 0 .63 I max
Physics 112, Spring 2010, Mar 3, Lecture 21
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LR Circuit: Disconnected From Battery
LR Circuit
Exponential decay.
A resistor and an inductor are connected in series to a battery. The time
constant for the circuit represents the time required for the current to reach:
1) 25% of the maximum current
2) 37% of the maximum current
2
3) 63% of the maximum current
4) 75% of the maximum current
LR Circuit
Switch at 2: the battery is disconnected from the circuit.
A resistor and an inductor are connected in series to a battery. The battery
is suddenly removed from the circuit. The time constant for of the circuit
represents the time required for the current to decrease to:
Large back EMF develops across L and prevents it dropping straight to zero.
I = I max e − t / τ
When t = τ :
τ =
L
R
e − t / τ = 0 .37
LR circuit similar to RC circuit (charging
and discharging of capacitor), but time
constant now is inversely proportional to R.
2) 37% of the original value
3) 63% of the original value
I = 0 .37 I max
Physics 112, Spring 2010, Mar 3, Lecture 21
1) 25% of the original value
4) 75% of the original value
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