Question
Two metal rings lie side-by-side on a table. Current in the left ring runs counterclockwise and is increasing with time. This induces a current in the right ring. This current runs
A) Clockwise B) Counterclockwise
when viewed from above
Inductance
Constant voltage – constant I, no curly electric field.
Increase voltage: dB/dt is not zero → emf NI For long solenoid: B = µ 0 d Change current at rate dI/dt: dΦ mag d ⎡ µ0 NI 2 ⎤ µ0N 2 dI emf1 = = πR = πR (one loop) dt dt ⎣⎢ d ⎦⎥ d dt µ N 2 dI emf = 0 πR2 d dt R emfbat emfcoil
ENC R emfbat emfcoil EC µ N 2 dI emf = 0 πR2 d dt Increasing I → increasing B dI dΦmag emfind = L emf = − dt µ N 2 dt L = 0 πR2 d R emf L – inductance, or self-inductance bat Unit: Henry = Volt.second/Ampere emfind ∆Vsol = emfind − rsol I
Increasing the current causes ENC to oppose this increase.
Inductance: Decrease Current
ENC
EC
R L emfbat
emfind dΦ dI mag emf = L emf = − ind dt dt
Conclusion: Inductance resists changes in current
Orientation of emfind depends on sign of dI/dt
2 Magnetic Field Energy Density Electric energy 1 = ε E 2 Volume 2 0 LI2 Is there energy stored in magnetic field?
2 2 dI 1 ⎛ µ0 N 2 ⎞⎛ Bd ⎞ P = I∆V = I(emf ) = I(L ) Energy = ⎜ πR ⎟⎜ ⎟ dt 2 ⎝ d ⎠⎝ µ0N ⎠ I f ∆Energy = Pdt = L IdI 1 ⎛ B2 ⎞ ∫ ∫ Energy = ⎜ πR2d ⎟ I ⎜ ⎟ i 2 µ I f ⎝ 0 ⎠ 1 2 ⎛ 1 2 ⎞ ∆Energy = LI = ∆⎜ LI 2 2 1 1 2 Ii ⎝ ⎠ Energy = B V 2 µ0 2 µ0 N 2 NI L = πR B = µ Magnetic energy 1 1 2 d 0 d = B Volume 2 µ0
Field Energy Density Electric energy 1 = ε E 2 Volume 2 0
Magnetic energy 1 1 = B2 Volume 2 µ0
Electric and magnetic field energy density:
Energy 1 2 1 1 2 = ε0E + B Volume 2 2 µ0
3 Current in RL Circuit
∆Vbattery + ∆Vresistor + ∆Vinductor = 0 dI emf − RI − L = 0 battery dt I(t) = a + bect
ct ct emfbattery − Ra − Rbe − Lbce = 0 emf R a = battery Rb = −Lbc c = − R L R emf − t I(t) = battery + be L R emf If t is very long: I(t = ∞) = battery R
Current in RL Circuit
R emf − t I(t) = battery + be L R If t is zero: I(0) = 0 emf I(0) = battery + b ⋅1 = 0 R emf b = − battery R Current in RL circuit: R emfbattery ⎡ − t ⎤ I(t) = ⎢1− e L ⎥ R ⎣ ⎦
4 Time Constant of an RL Circuit
Current in RL circuit: R emfbattery ⎡ − t ⎤ I(t) = ⎢1− e L ⎥ R ⎣ ⎦
Time constant: time in which exponential factor drops e times
R L t = 1 τ = L R
Current in an LC Circuit
∆Vcapacitor + ∆Vinductor = 0
Q dI dQ − L = 0 I = − C dt dt d 2Q Q + LC = 0 dt2 Q = a + bcos()ct a + bcos(ct)+ LC(− bc2 cos(ct)) = 0 a=0 1 c = LC ⎛ t ⎞ ⎛ t ⎞ Q = bcos⎜ ⎟ Q = Q0 cos⎜ ⎟ ⎝ LC ⎠ ⎝ LC ⎠
5 Current in an LC Circuit
⎛ t ⎞ Q = Q0 cos⎜ ⎟ ⎝ LC ⎠ dQ I = − dt
Current in an LC circuit Q ⎛ t ⎞ I = 0 sin⎜ LC ⎝ LC ⎠
Period: T = 2π LC Frequency: f = 1/( 2π LC )
Non-ideal LC Circuit
∆Vcapacitor + ∆Vinductor + ∆VR = 0
Q dI dQ − RI − L = 0 I = − C dt dt dQ d2 Q QR++ LC =0 dt dt 2
at QQe= 0 10++aR a2 LC = −±R RLC2 −4 a = 2LC
6 Energy in an LC Circuit Q2 Initial energy stored in a capacitor: 2C 2 Q0 At time t=0: QQQ=Q0 U = cap 2C π 1 At time t= LC : Q=0 U = LI 2 2 sol 2 1/4 of a period
System oscillates: energy is passed back and forth between electric and magnetic fields.
Energy in an LC Circuit
What is maximum current?
At time t=0: Q2 U = U +U = 0 total el mag 2C π At time t= LC : 2 1 U = U +U = LI 2 total el mag 2 max
2 1 2 Q0 Q0 LImax = I = 2 2C max LC
7 Energy in LC Circuit
U = Uelectric + Umagnetic (No dissipation in this circuit)
2 1 Q 1 2 dU d( ) d( LI ) Q dQ dI = 2 C + 2 = + LI = 0 dt dt dt C dt dt
dQ =−I As capacitor loses charge, current increases dt As capacitor gains charge, current decreases
Q dI − L = 0 C dt Same equation as obtained via considering potential differences
LC Circuit and Resonance
Frequency: f = 1/( 2π LC )
Radio receiver:
8 AC Current and a Coil
dΦ dB dI emf = −N coil ~ N ~ N 2 dt dt dt SlfidSelf induced emf opposes emf of an AC source making current smaller
AC source
If number of loops is very large there will be almost no
current in the circuit and emfind will be equal to emfAC of the AC source: dΦ emf = −emf = N coil AC ind dt
AC Current and a Coil: Add a Loop
dΦ emf = − coil loop dt dΦ emf = N coil AC dt
emf emf = − AC AC source loop N
9 Transformer
emf AC emfloop = − N prim
Nsec emfsec = − emf AC N prim
Energy conservation:
Nsec Isecemfsec = I primemf AC I prim = − Isec N prim
Faraday’s Law: Applications
Singl e h ome current : 100 A serv ice
∆Vwires=IRwires Transformer: emfHV IHV = emfhomeIhome Single home current in HV: <0.1 A Power loss in wires ~ I2
10 Recap: Inductance dΦ emf = − mag R dt emfbat emfcoil
µ N 2 dI emf = 0 πR2 d dt Increasing I → increasing B dI emf = L ind dt L – inductance, or self-inductance µ N 2 L = 0 πR2 d R Unit of inductance L: emfbat Henry = Volt.second/Ampere emfind Inductance resists changes in current
Recap: Current in RL Circuit
dI emf − RI − L = 0 battery dt
R emfbattery ⎡ − t ⎤ I(t) = ⎢1− e L ⎥ R ⎣ ⎦
Time constant : time in which exponential factor drops e times
R L t = 1 τ = L R
11 Recap: Current in LC Circuit d 2Q Q + LC = 0 dt 2 ⎛ t ⎞ Q = Q0 cos⎜ ⎟ ⎝ LC ⎠ dQ I = − dt Period: T = 2π LC Q0 ⎛ t ⎞ I(t) = sin⎜ ⎟ LC ⎝ LC ⎠ Frequency: f = 1/( 2π LC )
Non-ideal LC circuit:
∆Vcapacitor + ∆Vinductor + ∆VR = 0 Q dI − RI − L = 0 C dt
Recap: Energy in LC Circuit Q2 Initial energy stored in a capacitor: 2C 2 Q0 At time t=0: QQQ=Q0 U = cap 2C π 1 At time t= LC : Q=0 U = LI 2 2 sol 2 Q Maximum current: I = 0 max LC
System oscillates: Energy is passed back and forth between electric and magnetic fields.
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