-Self Inductance-Inductance of a Solenoid

-RL Circuit-Energy Stored in an Inductor

AP Physics C Mrs. Coyle

• Induced emf and induced current are caused by changing magnetic fields.

Self-InductanceSwitch Open: No current and no magnetic field

Switch Closed: current increases, creates an increasing magnetic field which in turn induces an induced emf

• The direction of the induced emf is opposite the direction of the emf of the battery

• Gradually the net current increases to an equilibrium value

• This effect is called self-inductance– Because the changing flux through the circuit and

the resultant induced emf arise from the circuit itself

• The emf εL is called a self-induced emf

Inductors• In general wires between resistors cause a small

self inductance which is ignored.

• However, some circuit elements such as solenoids can cause a significant inductance, L. These objects are called inductors.

• Symbol for an inductor:

Inductance• Induced emf:

• L is a proportionality constant called the inductance of the coil and it depends on the geometry of the coil and other physical characteristics

• The inductance is a measure of the opposition to a change in current

IL

dε L

dt I

LεLd dt

Inductance Units

• The SI unit of inductance is the henry (H)

• Named for Joseph Henry

AsV

1H1

Inductance in a Solenoid (Coil)

The polarity of the induced emf, from Lenz’s Law, opposes the change in magnetic flux.

Inductance of a Solenoid• Assume a uniformly wound solenoid having N

turns and length ℓ– Assume ℓ is much greater than the radius of the

solenoid

• The interior magnetic field is uniform:

I Io o

NB μ n μ

Inductance of a Solenoid

• Faraday’s Law

• Equate eq. 1 and 2 solve for L for a solenoid:

• Remember also :

I

BN

L

IL

dε L

dt

Bdε

dt

I LεL

d dt

(eq.1)

(eq.2)

Inductance of a Solenoid

• The magnetic flux through each turn is

• Note that L depends on the geometry of the object

IB o

NABA μ

I

BN

L

2oμ N A

LI Io o

NB μ n μ

RL Circuit (Resistor and Inductor)• Kirchhoff’s loop rule for:

0I

Id

ε R Ldt

1I Rt Lεe

R

L

R 1I t τε

eR

Time Constant,

RL Circuit Current

• The inductor affects the current exponentially• The current does not instantly increase to its

final equilibrium value• If there is no inductor, the exponential term

goes to zero and the current would instantaneously reach its maximum value as expected

1I Rt Lεe

R

RL Circuit, Time Constant

= L / R

When T=

I= 0.632 Ieq,

then, the current has reached 63.2% of its equilibrium value.

1I t τεe

R

I 11ε

eR

LR Circuit, Charging

dIRI L 0

dt

t /I 1 eR

L

R

Prove:

LR Circuits, discharging

dIRI L 0

dt

t /oI I e

L

R

Prove:

Energy Stored in an Inductor, U• The rate at which energy is being supplied

by the battery (Power=IV)2 I

I I Id

ε R Ldt

21U LI

2

0

II IU L d

II

dU dL

dt dt

Remember for a capacitor:21

U CV2

Ex: RL CircuitEx: RL Circuit

a) What happens JUST AFTER the switch is closed?

b) What happens LONG AFTER switch has been closed?

c) What happens in between?

Note: At t=0, a capacitor acts like a regular wire; an inductor acts like an open wire.

After a long time, a capacitor acts like an open wire, and an inductor acts like a regular wire.

Ex: RL CircuitsEx: RL Circuits

Immediately after the switch is closed, what is the potential difference across the inductor?(a) 0 V(b) 9 V(c) 0.9 V

• Immediately after the switch, current in circuit = 0.• So, potential difference across the resistor = 0• So, the potential difference across the inductor = E9 V

10

10 H9 V

Ex: RL CircuitsEx: RL Circuits• Immediately after the

switch is closed, what is the current i through the 10 resistor?(a) 0.375 A(b) 0.3 A(c) 0

• Long after the switch has been closed, what is the current in the 40 resistor?

(a) 0.375 A

(b) 0.3 A

(c) 0.075 A

40

10 H

3 V

10

• Immediately after switch is closed, current through inductor = 0.• Hence, current trhough battery and through 10 resistor is i = (3 V)/(10) = 0.3 A

• Long after switch is closed, potential across inductor = 0.

• Hence, current through 40 resistor = (3 V)/(40) = 0.075 A

Ex: RL CircuitsEx: RL Circuits

• How does the current in the circuit change with time?

0dt

diLiRE

tL

R

eR

Ei 1

“Time constant” of RL circuit = L/R

i

i(t) Small L/R

Large L/R

t

• The switch has been in position “a” for a long time.

• It is now moved to position “b” without breaking the circuit.

• What is the total energy dissipated by the resistor until the circuit reaches equilibrium?

• When switch has been in position “a” for long time, current through inductor = (9V)/(10) = 0.9A.

• Energy stored in inductor = (0.5)(10H)(0.9A)2 = 4.05 J• When inductor “discharges” through the resistor, all this stored

energy is dissipated as heat = 4.05 J.

9 V

10

10 H

Ex: Energy Stored in a B-FieldEx: Energy Stored in a B-Field

Energy Density of a Magnetic Field• U = ½ L I 2

• Aℓ is the volume of the solenoid

• Magnetic energy density, uB :

• This applies to any region in which a magnetic field exists (not just the solenoid)

2 221

2 2oo o

B BU μ n A A

μ n μ

2

2Bo

U Bu

A μ

2oμ N A

Lo

BI

N

Example 32.5: The Coaxial Cable

• Calculate L for the cable• The total flux is

• Therefore, L is

• The total energy is

ln2 2

I Ibo o

B a

μ μ bB dA dr

πr π a

ln2I

B oμ bL

π a

221

ln2 4

II oμ b

U Lπ a

Prob.#3

A 2.00-H inductor carries a steady current of 0.500 A. When the switch in the circuit is opened, the current is effectively zero after 10.0 ms. What is the average induced emf in the inductor during this time?

Ans: 100V

Prob#5 A 10.0-mH inductor carries a current I = Imax

sin ωt, with Imax = 5.00 A and ω/2π = 60.0 Hz. What is the back emf as a function of time?

Ans: (18.8V)cos (377t)

Prob.# 7

An inductor in the form of a solenoid contains 420 turns, is 16.0 cm in length, and has a cross-sectional area of 3.00 cm2. What uniform rate of decrease of current through the inductor induces an emf of 175 μV?

Ans: -0.421A/s

Prob.#14

Calculate the resistance in an RL circuit in which L = 2.50 H and the current increases to 90.0% of its final value in 3.00 s.

Prob.#16

dIRI L 0

dt

t /oI I e Show that is a solution

of the differential equation

where and Io is the current

at t=0.

L

R

Prob. #20

A 12.0-V battery is connected in series with a resistor and an inductor. The circuit has a time constant of 500 μs, and the maximum current is 200 mA. What is the value of the inductance?

Prob.# 24

A series RL circuit with L = 3.00 H and a series RC circuit with C = 3.00 μF have equal time constants. If the two circuits contain the same resistance R, (a) what is the value of R and (b) what is the time constant?

Prob.#26

• A) What is the current in the circuit a long time after the switch has been in position a?

• B) Now the switch is thrown from a to b. Compare the initial voltage across each resistor and across the inductor.

• C) How much time elapses before the voltage across the inductor drops to 12.0V?

12.0

2.00 H

12 V

1200

Prob.#31

An air-core solenoid with 68 turns is 8.00 cm long and has a diameter of 1.20 cm. How much energy is stored in its magnetic field when it carries a current of 0.770 A?

Prob.#33

On a clear day at a certain location, a 100-V/m vertical electric field exists near the Earth’s surface. At the same place, the Earth’s magnetic field has a magnitude of 0.500 × 10–4 T. Compute the energy densities of the two fields.

Prob.# 36

A 10.0-V battery, a 5.00-Ω resistor, and a 10.0-H inductor are connected in series. After the current in the circuit has reached its maximum value, calculate (a) the power being supplied by the battery, (b) the power being delivered to the resistor, (c) the power being delivered to the inductor, and (d) the energy stored in the magnetic field of the inductor.