Download 1 Inductance, RL Circuits, Energy Stored in an Inductor

-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-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 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:
εL
dI
L
εL   L
d I dt
dt
• 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
Inductance Units
• The SI unit of inductance is
the henry (H)
V s
1H  1
A
• Named for Joseph Henry
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:
B  μo n I  μo
N
I
Inductance of a Solenoid
dI
εL   L
dt
dB
ε
dt
• Faraday’s Law
• Equate eq. 1 and 2 solve for L
for a solenoid:
NB
L
• Remember also :
I
εL
L
d I dt
(eq.1)
(eq.2)
Inductance of a Solenoid
NB
L
I
• The magnetic flux through each turn is
 B  BA  μo
NA
B  μo n I  μo
N
I
I
L
μo N A
2
• Note that L depends on the geometry of the
object
RL Circuit (Resistor and Inductor)
• Kirchhoff’s loop rule for:
dI
ε IR L
0
dt
Time Constant, 
L

R
ε
Rt L
I  1  e

R
ε
t τ
I
1 e
R


RL Circuit Current
ε
Rt L
I  1  e

R
• 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
RL Circuit, Time Constant ,
I
=L/R
When T=,
ε
I
1 e 1
R

ε
1  e t τ 

R

I= 0.632 Ieq,
then, the current has reached 63.2% of its equilibrium value.
LR Circuit, Charging
dI
  RI  L  0
dt
Prove:
I

R
t / 
1

e


L

R
LR Circuits, discharging
dI
RI  L  0
dt
Prove:
I  Io e
L

R
t / 
Energy Stored in an Inductor, U
• The rate at which energy is being supplied
by the battery (Power=IV)
dI
Iε  I RLI
dt
2
dU
dI
LI
dt
dt
I
U  L I d I
0
Remember for a capacitor:
1
U  CV 2
2
1 2
U  LI
2
Ex: 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 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
10 W
9V
• 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 H
Ex: RL Circuits
• Immediately after the
3V
switch is closed, what is
the current i through the
10 W resistor?
(a) 0.375 A
(b) 0.3 A
• Immediately after switch is closed,
(c) 0
current through inductor = 0.
40 W
10 W
10 H
• Hence, current trhough battery and
through 10 W resistor is
i = (3 V)/(10W) = 0.3 A
• Long after the switch has been closed, what
is the current in the 40W resistor?
• Long after switch is closed, potential
(a) 0.375 A
across inductor = 0.
(b) 0.3 A
• Hence, current through 40W resistor
= (3 V)/(40W) = 0.075 A
(c) 0.075 A
Ex: RL Circuits
• How does the current
in the circuit change
with time?
i
di
E  iR  L  0
dt
R

t

E
L
i  1  e 
R

“Time constant” of RL circuit = L/R
i(t)
Small L/R
Large L/R
t
Ex: Energy Stored in a B-Field
• 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?
10 W
9V
10 H
• 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.
Energy Density of a Magnetic Field
• U=½LI
2
L
μo N A
2
B
I
o N
2
2


1
B
B
U  μo n 2 A 
A
 
2
2 μo
 μo n 
• Aℓ is the volume of the solenoid
• Magnetic energy density, uB :
U
B2
uB 

A
2μo
• This applies to any region in which a magnetic field
exists (not just the solenoid)
Example 32.5: The Coaxial Cable
• Calculate L for the cable
• The total flux is
 B   B dA  
b
a
μo I
μo I
b
dr 
ln  
2πr
2π
a
• Therefore, L is
 B μo
b
L

ln  
I
2π  a 
• The total energy is
1 2 μo I 2  b 
U  LI 
ln  
2
4π
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.
Show that
Prob.#16
 t /  is a solution
I  Io e
dI
0
of the differential equation RI  L
dt
where
at t=0.
L

R
and Io is the current
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
12.0 W
12 V
1200 W
2.00 H
• 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?
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 10.0H 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.