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 = ½LI2.
Energy density of magnetic field filling volume: u = ½B2/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 & 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
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Example: Calculating M
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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 = ½CV2 = ½Q2/C.
Uinductor = U = ½LI2.
ULC = (½Q02/C) cos2(ωt+φ) +
(½L) (ωQ0)2 sin2(ωt+φ)
ULC = (½Q02/C) [cos2(ωt+φ) +
sin2(ωt+φ)] = (½Q02/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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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 (< 4L/C): damped oscillation
R2 = 4L/C: ω’ = 0, critical damping Q → A’ e-(R/2L)t
R2 > 4L/C: ω’2 < 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