Energy of an Inductor• How much energy is stored

in an inductor when a current is flowing through it?

R

ε

a

b

L

I I

• Start with loop rule:dtdILIR +=ε

• From this equation, we can identify PL, the rate at which energy is being stored in the inductor:

dtdILI

dtdUPL ==

• We can integrate this equation to find an expression for U, the energy stored in the inductor when the current = I:

⇒ 0 0

= =∫ ∫U I

U dU LI dI 2

21 LIU =

dtdILIRIεI += 2

• Multiply this equation by I:

Magnetic field Energy in a toroid

Consider a toroid magnet, the B fieldis, B = μ0NI/2πr. The energy is,

Substituting the B field into theEqn., we have,

22

0222

121 I

rANLIU ⎟

⎟⎠

⎞⎜⎜⎝

⎛==

πμ

( ) ( ) 0

2

2

2220

02

220

2221

221

2 μπ

μμπ

μπ

Br

INr

INrA

U=⎟

⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

densityEnergyBvolume

UrA

U===

0

2

22 μπ

Approximates B by value atcenter of coil.

Energy in Electric Fields and Magnetic FieldsIn chapter 24.3, we discussed energy in a parallel plate with area A and separation d, The electric field energy in the capacitor was

( ) ( ) ( ) 20

202221

21

21

21 EAdEd

dAEdCCVU εε

=⎟⎠⎞

⎜⎝⎛===

202

Evolumeenergy

AdU ε

==

Now we find the magnetic field energy in the toroid magnet is

2

021

2B

volumeenergy

rAU

μπ==

The fields are proportional to the energy density22, EBrr

Inductors in CircuitsGeneral rule: inductors

resist change in current:

ε = - L di/dt

• Hooked to current source

– Initially, the inductor behaves like an open switch.

– After a long time, the inductor behaves like an ideal wire.

• Disconnected from current source

– Initially, the inductor behaves like a current source.

– After a long time, the inductor behaves like an open switch.

RI

ε

a

b

L

I

Lecture 17, Act 2• At t=0 the switch is thrown from position

b to position a in the circuit shown: – What is the value of the current I∞ a

long time after the switch is thrown?

(a) I∞ = 0 (b) I∞ = ε/2R (c) I∞ = 2ε/R

(a) I0 = 0 (b) I0 = ε/2R (c) I0 = 2ε/R

2A

– What is the value of the current I0 immediately after the switch is thrown?

2B

R

a

b

ε

R

L

II

(a) I0 = 0 (b) I0 = ε/2R (c) I0 = 2ε/R

–What is the value of the current I0 just after the switch is opened?

2C

• After a long time the switch is opened.

Lecture 17, Act 2

(a) I∞ = 0 (b) I∞ = ε/2R (c) I∞ = 2ε/R

2A

• A long time after the switch is thrown, the current approaches an asymptotic value: as t → ∞, dI/dt → 0.

• As dI/dt → 0, the voltage across the inductor → 0. Therefore,I∞ = ε/2R.

• At t=0 the switch is thrown from positionb to position a in the circuit shown: –What is the value of the current I∞ a

long time after the switch is thrown?

a

b

ε

R

L

II

R

Lecture 17, Act 2• At t=0 the switch is thrown from position

b to position a in the circuit shown: –What is the value of the current I0

immediately after the switch is thrown?(a) I0 = 0 (b) I0 = ε/2R (c) I0 = 2ε/R

2B

• Just after the switch is thrown, the rate of change of current is as large as it can be (we had been assuming it was ∞!)

• The inductor limits dI/dt to be initially equal to ε/L. The voltage across the inductor = ε; the current, then, must be 0!

• Another way: the moment the switch is thrown, the current tries to generate a huge B-field. There is a huge change in flux through coil—an emf is generated to oppose this. Initially, then, no current flows through no voltage drop across the resistors.

a

b

ε

R

L

II

Lecture 17, Act 2

(a) I0 = 0 (b) I0 = ε/2R (c) I0 = 2ε/R

2C

• Just after the switch is thrown, the inductor induces an emf to keep current flowing: emf = L dI/dt (can be much larger than ε)

• However, now there’s no place for the current to go charges build up on switch contacts high voltage across switch gap

•If the electric field exceeds the “dielectric strength”(~30 kV/cm in air) breakdown SPARK!

This phenomenon is used in “flyback generators” to create high voltages; it also destroys lots of electronic equipment!

a

b

ε

R

L

II

• After a long time the switch is opened.

–What is the value of the current I0just after the switch is opened?

RL Circuits, Quantitative

• At t=0, the switch is closed and the current I starts to flow.

Note that this equation is identical in form to that for the RC circuit with the following substitutions:

RI

ε

a

b

L

I

• Loop rule:

0dIIR Ld t

ε − − =

⇒ Therefore, RCRC =τRL

RL =τ

⇒ RC: 0Q dQε RC dt

− − = LR→

RC

→1

IQ →

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−RCt

eCQ 1ε

• To find the current I as a function of time t, we need to choose an exponential solution which satisfies the boundary condition:

⇒

RL Circuits

0)( =∞=tdtdI

RtI ε

=∞= )(

• We therefore write: ( )LRteRεI /1 −−=

• The voltage drop across the inductor is given by:LRt

L εedtdILV /−==

R

τRL = LRε

a

b

L

I I

RL Circuit (ε on)

Current

Max = ε/R

63% Max at t=L/R

( )LRteR

I /1 −−=ε

L/R

t

I

2L/R

0

ε/R

VL

0t

εVoltage on L

Max = ε/R

37% Max at t=L/R

LRtL e

dtdILV /−== ε

Sketch curves !

RL Circuits• After the switch has been in

position a for a long time, redefined to be t=0, it is moved to position b.

R

ε

a

b

L

I I

• Loop rule:

0=+dtdILIR

• The appropriate initial condition is:R

tI ε== )0(

• The solution then must have the form:

LRteR

I /−=ε

LRtL e

dtdILV /−−== ε

RL Circuit (ε off)

0

-ε

VL

t

L/R

t

2L/R

I

0

ε/RCurrent

Max = ε/R

37% Max at t=L/R

LRteR

I /−=ε

Voltage on L

Max = -ε

37% Max at t=L/R

LRtL e

dtdILV /−−== ε

ε on ε off

t0

-ε

I

t

0

ε/R L/R 2L/R

LRteR

I /−=ε

LRtL e

dtdILV /−−== ε

VL

L/R

t

2L/R

0

ε/R

I

0t

ε

( )LRteR

I /1 −−=ε

LRtL e

dtdILV /−== ε

VL

Lecture 17, Act 3• At t=0, the switch is thrown from

position b to position a as shown:– Let tI be the time for circuit I to

reach 1/2 of its asymptotic current.– Let tII be the time for circuit II to

reach 1/2 of its asymptotic current.– What is the relation between tI and

tII?(a) tII < tI (b) tII = tI (c) tII > tIε

a

b L

I I

R

L

II

IR

ε

a

b

L

I

R

I

Lecture 17, Act 3

• We must determine the time constants of the two circuits by writing down the loop equations.

0d II R L I Rd t

ε − − − =

0dI d IL IR Ld t d t

ε − − − =

I:

II:

RL

I 2=τ

RL

II2

=τThis confirms that

inductors in series add!

(a) tII < tI (b) tII = tI (c) tII > tI

• At t=0, the switch is thrown from position b to position a as shown:– Let tI be the time for circuit I to

reach 1/2 of its asymptotic current.– Let tII be the time for circuit II to

reach 1/2 of its asymptotic current.– What is the relation between tI and

tII?ε

a

b L

I I

R

L

II

IR

ε

a

b

L

I

R

I

7) At t = 10 hrs the switch is opened, abruptly disconnecting the battery from the circuit. What will happen to all the energy stored in the solenoid?

Preflight 17:

Energy stored in the inductor: U = 1/2 L I2

An inductor doesn’t like change!!!

When the switch is opened, this energy is dissipated in the resistor.

When the switch is opened, the inductor will try to maintain the current that was flowing through it before the switch is opened. Since the battery is disconnected from the circuit, the energy which is necessary to keep current flowing through the resistor is provided by theinductor.