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INDUCTORS AND INDUCTANCE
An inductor (symbol ) can be used to produce a desired magnetic field. If we establish a current i in the windings (turns) of the solenoid which can be treated as our inductor, the current produces a magnetic flux ΦB through the central region of the inductor. The inductance of the inductor is then The SI unit of inductance is the tesla–square meter per ampere (T m2/A). We call this the henry (H), after American physicist Joseph Henry,
INDUCTORS
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L =NΦB
i N = nl ΦB = BANΦB = nlBA B = µoin
L =nl( ) µoin( )(A)
iL = µon
2lALl
= µon2A
Unit of inductance
Tm2/A = henry (H)
Symbol:
INDUCTANCE
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NΦB = Li
εL = −d (NΦB )
dtεL = Li
Loop with current - magnetic flux through loop
If you change the current, you change the flux
An emf is induced that opposes the change in the original current (only occurs while the current is changing)
Self-Inductance - the induction of a voltage in a current-carrying wire when
the current in the wire itself is changing. In the case of self-inductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit.
In a circuit containing a resistor and an inductor, the current increases as shown to reach its steady-state value:
The RL circuit shown has a resistor and an inductor connected in series. A constant voltage V is applied when the switch is closed. The (variable) voltage across the resistor is given by: V = iR The (variable) voltage across the inductor is given by: V = L(di/dt) Kirchhoff's voltage law says that the directed sum of the voltages around a circuit must be zero. This results in the following differential equation: Ri + L(di/dt) = V Once the switch is closed, the current in the circuit is not constant. Instead, it will build up from zero to a steady state. Solving the Differential Equation for a Series RL Circuit: The solution of this differential equation is:
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i =VR(1− e−(R /L )t )
energizing
deenergizing
1. An RL circuit has an emf of 5 V, a resistance of 50 Ω, an inductance of 1 H, and no initial current.
Write an expression for the current in the circuit at any time t. Distinguish between the transient and steady-state current. Calculate a numerical value for the time constant.
EXAM
PLE
SOLU
TION
2. A series RL circuit with R = 50 Ω and L = 10 H has a constant voltage V = 100 V applied at t = 0 by the closing of a switch. Write or calculate as indicated: (a) the equation for i (b) the current at t = 0.5 s (c) the expressions for VR and VL (d) the time at which VR = VL
EXAM
PLE
Energy Stored in a Magnetic Field
Recall that electric energy is stored between the plates of a capacitor. Likewise, magnetic energy is stored in the magnetic field of an inductor. A short time after the switch in this circuit is closed, Kirchoff’s loop rule gives:
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εo − iR − Ldidt
= 0
εoi = i 2R + Li didt
The final term is the rate at which electric potential energy is supplied to the inductor, where it is stored:
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dUmdt
= Li didt
dUm = LididUm∫ = Lidi∫
Um =12Li 2 +C
Assuming that the inductor is a solenoid:
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B = µoni
L = µon2Al
Um =12Li 2 =
12
µon2Al B
µon
"
# $
%
& '
2
=B2
2µoAl
Al is the volume of space within the solenoid. The energy per unit volume is the energy density.
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um =B2
2µo
Example, Energy stored in a magnetic field:
Current in Coil 1 produces a magnetic flux through Coil 2. If the current in Coil 1 is changed, an emf is induced in Coil 2 and also in Coil 1.
MUTUAL INDUCTION
MUTUAL INDUCTION
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L =NΦi
M 21 =N 2Φ21i1
M 21i1 = N 2Φ21
M 21di1dt
=N 2dΦ21dt
−M 21di1dt
= ε2
−M12di2dt
= ε1
M 21 = M12 = M
ε2 = −M di1dt
ε1 = −M di2dt
Mag. flux in 2 due to current in 1
Induced emf in 2 due to di/dt in 1
Flux change in 2 due to di/dt in 1
Exchange roles of the two coils (battery moved to coil 2)