Download March 18 Induction and Inductance Chapter 31

March 18
Induction and
Inductance
Chapter 31
Review - Self Inductance
>
Self-induce emf, EL appears
in any coil in which the
current is changing
di
εL = −L
dt
>
€
Direction of EL follows
Lenz’s law and opposes the
change in current
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Review - Mutual Inductance
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What is induced emf in coil
1 from a changing current
in coil 2?
di2
ε1 = −M
dt
where
N1Φ12 N 2Φ 21
M=
=
i2
i1
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Review: RC circuit
>
RC circuit is a resistor
and capacitor in series
Charging up a capacitor
(switch at a)
o Kirchhoff rule for loop:
o
dq
q
R+ −E = 0
dt
C
(=differential equation)
o Solution
€
q = CE (1−e
March 18, 2004
−t τ c
o
q = q0 e
)
τ C184=
where PHY
Discharging capacitor
(switch at b)
RC
−t τ c
4
Inductance
>
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RL circuit is a resistor
and inductor in series
Close switch to point a
o
Initially i is increasing
through inductor so EL
opposes rise and i through
R will be
i <ε R
o
ε
L
di
= −L
dt
Long time later, i is
constant so EL =0 and i in
circuit is
i =ε R
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RL Circuit - Differential Equation
>
>
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Initially an inductor acts to
oppose changes in current
through it
Long time later inductor acts
like ordinary conducting wire
Apply Kirchhoff loop rule right
after switch has been closed at
a
ε
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di
− iR − L = 0
dt
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RL Circuit Solution
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Differential equation .…… similar to RC circuit:
dq
q
R+ −E = 0
dt
C
di
L + iR−ε = 0
dt
>
Solution is
ε
−t τ L
(
i = 1− e )
R
€
>
Inductive time constant is
L
τL =
R
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>
PHY 184
Satisfies conditions:
o
At t=0, i = 0
o
At t=∞, i =E/R
7
“Discharging”=Stop Current
>
>
Now move switch to position b
so battery is out of system
Current will decrease with
time and loop rule gives
di
iR + L
=0
dt
>
Solution is
>
ε −t τ L
−t τ L
i= e
= i0e
R
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Satisfies conditions
o At t=0, i =i0 =E/R
o
At t=∞, i = 0
8
RL circuits Summary
o
Circuit is closed (switch to “a”)
E
i=
1−e−t τ L
R
(
o
)
Circuit is opened (switch to “b”)
E −t τ L
−t τ L
i= e
= i0e
R
• Time constant is
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Switch at a, current through
inductor is:
> Initially i = 0 (acts like
broken wire)
L
τL =
R
>
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Long time later i = E/R (acts
like simple wire)
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Problem 31-5
>
>
>
>
>
Have a circuit with resistors
and inductors
What is the current through
the battery just after closing
the switch?
Inductor oppose change in
current through it
Right after switch is closed,
current through inductor is 0
Inductor acts like broken wire
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Problem 31-5, continued
>
Apply loop rule
E −iR = 0
>
Immediately after switch
closed, current through the
battery is
E
i=
R
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Problem 31-5, continued
>
>
>
What is the current through the
battery a long time after the
switch has been closed?
Currents in circuit have
reached equilibrium so inductor
acts like simple wire
Circuit is 3 resistors in parallel
E
i=
Req
R
Req =
3
Remember the “water slides”
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Energy
>
>
How much energy is stored
in a B field?
Conservation of energy
expressed in loop rule
ε
>
di
= L + iR
dt
>
Multiply each side by i
di 2
εi = Li + i R
dt
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P=iE is the rate at which
the battery delivers energy
to rest of circuit
P=i 2 R is the rate at which
energy appears as thermal
energy in resistor
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Energy (2)
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Middle term is rate at
which energy dUB /dt is
stored in the B field
>
Energy stored in magnetic
field
€
>
Similar to energy stored
in electric field
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dU B
di
= Li
dt
dt
1 2
U B = Li
2
2
1q
UE =
2C
14
Energy Density
>
>
What is the energy
density of a B field?
Energy density, uB is
energy per unit
volume
UB
uB =
Al
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Magnetic energy density
2
1B
uB =
2 µ0
>
Similar to electric energy
density
1
2
uE = ε 0 E
2
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Where does this come from?
B = µ0 ni
2
L = lµ0 n A ⇒
1 2
2
1
2
2
2 2
1
2
2 2
2
0
U€
B = i L = lµ0 n i A = Al(n i µ )/ µ0
U B / Al = 12 B / µ0
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