1 apc self inductance

7/22/2019 1 APC Self Inductance

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-Self Inductance

-Inductance of a Solenoid

-RL Circuit-Energy Stored in an Inductor

AP Physics C

Mrs. Coyle

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• Induced emf and induced current are causedby changing magnetic fields.



Self-Inductance

Switch 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 equilibriumvalue



• 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 theinductance of the coil and it depends on thegeometry of the coil and other physical

characteristics• The inductance is a measure of the opposition

to a change in current

I L

d ε L

dt I

Lε

Ld dt



Inductance Units

• The SI unit of inductance is

the henry (H)

• Named for Joseph Henry

AsV1H1



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 I o o

N

B μ n μ




• Faraday’s Law

• Equate eq. 1 and 2 solve for L

for a solenoid:

• Remember also : I

B

N L

I

L

d

ε L dt

Bd

ε dt

I L

ε L

d dt

(eq.1)

(eq.2)




• The magnetic flux through each turn is

• Note that L depends on the geometry of theobject

I B o

NA

BA μ

I

B

N

L

2

o μ N A

L I I o o

N B μ n μ



RL Circuit (Resistor and Inductor)

• Kirchhoff’s loop rule for:

0 I

I d

ε R Ldt

1 I

Rt Lε e

R

L

R

1 I t τ ε

e

R

Time Constant,



RL Circuit Current

• The inductor affects the current exponentially

• The current does not instantly increase to itsfinal equilibrium value

• If there is no inductor, the exponential termgoes to zero and the current wouldinstantaneously reach its maximum value as

expected

1 I Rt Lε

eR



RL Circuit, Time Constant ,

= L / R

When T=,

I= 0.632 Ieq,

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

1 I t τ ε

eR

I 1

1

ε e

R

LR Ci i Ch i



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 I d

ε R Ldt

21

U LI20

I

I I U L d

I I

dU d L

dt dt

Remember for a capacitor:21

U CV

2



Ex: RL Circuit

a) What happens JUST AFTERthe switch is closed?

b) What happens LONG AFTERswitch 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 Circuits

Immediately after the switch isclosed, what is the potentialdifference 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 = 9 V

10 W

10 H9 V



Ex: RL Circuits

• Immediately after theswitch is closed, what is

the current i through the10 W resistor?

(a) 0.375 A

(b) 0.3 A

(c) 0

• Long after the switch has been closed, what

is the current in the 40W resistor?

(a) 0.375 A

(b) 0.3 A

(c) 0.075 A

40 W

10 H

3 V

10 W

• Immediately after switch is closed,

current through inductor = 0.• Hence, current trhough battery and

through 10 W 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)/(40W) = 0.075 A



Ex: RL Circuits

• How does the currentin the circuit changewith time?

0 dt

di

LiR E

t L

R

e

R

E i 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 energydissipated by the resistor until

the circuit reaches equilibrium?

• When switch has been in position “a” for long time, current

through inductor = (9V)/(10W) = 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 W

10 H

Ex: 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 fieldexists (not just the solenoid)

22

21

2 2o

o o

B BU μ n A A

μ n μ

2

2B

o

U Bu

A μ

2

o μ N A

Lo

B

I 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 I b o o

Ba

μ μ bB dA dr πr π a

ln2 I

B o μ bLπ a

2

21ln

2 4

I I

o μ 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 inthe inductor during this time?

Ans: 100V



Prob#5

A 10.0-mH inductor carries a current I = Imaxsin 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 throughthe 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 theinductance?



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 longtime after the switch has been in position a?

• B) Now the switch is thrown from a to b.

Compare the initial voltage across eachresistor and across the inductor.

• C) How much time elapses before the voltage

across the inductor drops to 12.0V?

12.0 W

2.00 H

12 V

1200 W



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 suppliedby the battery, (b) the power being delivered

to the resistor, (c) the power being delivered

to the inductor, and (d) the energy stored inthe magnetic field of the inductor.

1 apc self inductance

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