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)