# Physics 227: Lecture 21 Mutual Inductance, LC Circuits

advertisement ```Physics 227: Lecture 21
Mutual Inductance, LC Circuits
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Lecture 20 review:
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Self-inductance of coil of wire: ξinduced = -N dΦB/dt = -L di/dt,
with L = NΦB/i.
Solenoid: L = μ0n2V with n the turns/unit length and V the
volume.
RL circuit time constant τ = L/R.
Energy stored in an inductor: U = &frac12;LI2.
Energy density of magnetic field filling volume: u = &frac12;B2/μ0.
Monday, November 28, 2011
Exam 2 Results
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Average: 60%, same as exam 1
Easiest problems: RC circuit time constant(90%), radius formula
for safe wire (88%), drift speed formula of electrons in wire
(84%), ...
Hardest problems: energy change when current carrying coil in B
field flipped 180o (22%), field between two infinite current sheets
(22%), relative magnitude of electric and magnetic forces
between two equal charges (37%), ...
Monday, November 28, 2011
Review RL iClicker
The switch S1 is closed so that a current
flows through an inductor of inductance L
and a resistor of resistance R as shown.
The final value of the current with S1
closed and S2 open is...
A. proportional to RL.
B. proportional to a ratio of R and L.
C. proportional to 1/LR.
D. independent of L.
E. 0.
Monday, November 28, 2011
Mutual Inductance
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The changing field in a coil leads to a
self-inductance of: ξinduced = -N dΦB/dt
= -L di/dt, with L = NΦB/i.
Changing external magnetic fields lead
to emf as well, as we know:
ξ2 = -N2 dΦB2/dt.
If the flux ΦB2 arises from an
electromagnet with current i1, then we
can write N2 ΦB2 = M21 i1, where M21 is
the mutual inductance. M21, like L, is a
constant that depends on geometry
and number of coils.
This leads to ξ2 = -N2 dΦB2/dt = -M21
di1/dt.
Monday, November 28, 2011
Mutual Inductance
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If the analysis is redone for the
induced emf in loop 1 and the current
in loop 2, it it is found that M21 =
N2ΦB2/i1 = M12 = N1ΦB1/i2 ≡ M.
Thus ξ2 = -M di1/dt, and ξ1 = -M di2/dt.
Monday, November 28, 2011
Applications &amp; Issues
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Applications:
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Cordless power
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Unintended cross talk in
circuits
Transformers
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Issues
Monday, November 28, 2011
Example: Calculating M
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Monday, November 28, 2011
Field from loop 1: B1 = μ0N1I1/L = μ0n1I1.
ΦB2 = AB1 = μ0N1I1A/L.
M = N2ΦB2/I1 = μ0N1N2A/L.
LC Circuit Math
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Monday, November 28, 2011
Kirchoff’s Voltage Law:
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Capacitor: Q = CV → V = Q/C
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I = dQ/dt
Inductor: V = L dI/dt
KVL → Q/C + L dI/dt = 0
Q/C + L d2Q/dt2 = 0
d2Q/dt2 + Q/LC = 0
This should look familiar. Recall the
solution to d2x/dt2 + ω2x = 0 is
x = A cos(ωt+φ) - simple harmonic
motion with a frequency ω2 = 1/LC.
Thus: Q = Q0 cos(ωt+φ), and
I = -ωQ0 sin(ωt+φ)
Energy in an LC Circuit
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Monday, November 28, 2011
Thus: Q = Q0 cos(ωt+φ), and
I = -ωQ0 sin(ωt+φ)
Ucapacitor = U = &frac12;CV2 = &frac12;Q2/C.
Uinductor = U = &frac12;LI2.
ULC = (&frac12;Q02/C) cos2(ωt+φ) +
(&frac12;L) (ωQ0)2 sin2(ωt+φ)
ULC = (&frac12;Q02/C) [cos2(ωt+φ) +
sin2(ωt+φ)] = (&frac12;Q02/C)
The energy in the circuit is constant
- there is no resistor to dissipate it.
It shifts between the capacitor and
the inductor.
LC Circuits - Inductor + Capacitor
Monday, November 28, 2011
LRC Series Circuit
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Monday, November 28, 2011
Charge up the capacitor with the
emf.
Flip the switch and discharge the
capacitor through the resistor and
inductor.
What happens?
Qualitatively the same as with
pendulum with friction, etc.: a
damped harmonic oscillator.
There are oscillations in the current
of decreasing magnitude.
It is good to have some intuition
about the answer before doing the
math - we have it from LC before.
RLC Ciruit
Monday, November 28, 2011
RLC Circuit Math
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Monday, November 28, 2011
Kirchoff’s Voltage Law →
Q/C + L dI/dt + IR = 0
d2Q/dt2 + (R/L) dQ/dt + Q/LC = 0
This is the damped harmonic
oscillator with damping proportional
to the ``velocity’’: the solution is
Q = A e-(R/2L)t cos(ω’t+φ) - simple
harmonic motion with a frequency
ω’2 = (1/LC) - (R2/4L2).
R small (&lt; 4L/C): damped oscillation
R2 = 4L/C: ω’ = 0, critical damping Q → A’ e-(R/2L)t
R2 &gt; 4L/C: ω’2 &lt; 0, overdamped
LC iClicker
An inductor of inductance L and a capacitor
of capacitance C are connected as shown. The
value of the charge q on the capacitor
oscillates between positive and negative.
The potential difference between the
capacitor plates is...
A. proportional to q.
B. proportional to dq/dt.
C. proportional to d2q/dt2.
D. both A. and C.
E. all of A., B., and C.
Monday, November 28, 2011
LC iClicker
An inductor of inductance L and a capacitor
of capacitance C are connected as shown. The
inductance L and capacitance C are both
doubled.
What is the change in the time for an
oscillation cycle in the circuit?
A. 4x longer.
B. 2x longer.
C. Unchanged.
D. 1/2 as long.
E. 1/4 as long.
Monday, November 28, 2011
Have a nice Holiday.
See you Monday after
Thanksgiving.
Monday, November 28, 2011
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