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Transcript
Self Inductance
Self Inductance
A variable power supply is connected to a loop.
The current in the loop creates a magnetic field.
What happens when the power supply dial is turned
down reducing the current, or turned up increasing the
current?
Self Induction
R
The current does not go from zero to ε/R in the circuit
immediately after the switch is closed:
1. as the current flows through, magnetic flux through
the loop is set up
2. this is opposed by induced emf in the loop which
opposes the change in net magnetic flux
3. by Lentz’s law, the induced E-field opposes the
current flow
Self Induction
R
As a time-varying current flow through the conductor,
the same thing happens:
1. as the changing current flows through, the magnetic
flux through the loop changes
2. this is opposed by induced emf in the loop which
opposes the change in net magnetic flux
3. by Lentz’s law, the induced E-field opposes the
current flow
Demo
“Self-Inductance in a closed loop”
We must keep the shape and size of the loop fixed.
BI
(From the Biot-Savart Law)
 BI
d
dI
 

dt
dt
B
B
L
The self-inductance L is the proportionality constant. It
depends on the shape of the loop, that is, its geometry.
The self-induced ems is proportional to the rate of
change of the current:
dI
ε  L
dt
L
(definition of L)
Unit for L: 1 henry (H) = 1 (V•s)/A
From Faraday’s Law for N loops and our definition
of inductance, L:
dI
d
  L   N
dt
dt
L
An equivalent definition of L uses the integral of the
above where  is the flux through each loop produced
by current I in each loop:
LI  N 
To find L we use:

LN
I
Quiz
A flat circular coil with 10 turns (each loop
identical) has inductance L1 . A second coil, of
the same size, shape and current passing through
the conductor but with 20 turns, would have
inductance:
A) 2 L1
B) ½ L1
C) 4 L1
D) ¼ L1
E) L1
Also:
dI
  L
dt
L  

dI / dt
We can see that inductance is the measure of the
opposition to the change in current, I.
(Recall that resistance, R, is the opposition to current, I:
R=V/I)
Example 1: Long solenoid
B
2r
I
I
l
Given:
N = 300 turns
r = 1 cm
l = 25cm
Calculate L
Solution
Example 2: Long solenoid
B
2r
I
I
l
Given:
N = 300 turns
r = 1 cm
l = 25cm
dI/dt = -50 A/s
Calculate the self-induced emf
Solution
Example 3: Self-Inductance of a coaxial cable

b
a
r
In the gap (a < r < b):
I
o I
B
2r

Show that
 
o b
L
ln
a
2
Solution
Kirchhoff’s Loop Rule terms (again)
Voltage change in going along path from left to right:
- resistor,
V   IR
- capacitor,
q
V 
- inductor,
dI
V   L
dt
I
q
q
C
I
Path direction
same as current
Example 4
What is the current 2 ms
after the switch is closed?
Hint: write down Kirchoffs
loop rule and solve the
differential equation for I
12 V
20mH
Energy
How much work is done
by an external emf to
increase the current
from 0 to Ifinal ?
I

dI
Kirchoff’s law gives:   L  0
dt
(Recall, P=VI)
+
εL=  LdI/dt
-
dI
 I  LI
dt
Power supplied by external emf
=
power absorbed by inductor
dU
dI
 LI
dt
dt
Power absorbed by inductor is:
L
dU  LIdI
L
The total potential energy stored is:
If
U  L  I dI  LI
1
2
L
2
f
0
U L  LI
1
2
2
Find the potential energy of an inductor with L=400mH and
a final current of 10A.