Self Inductance Consider a solenoid with n turns/m, length l, current i, and cross sectional area A....
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Self Inductance r a solenoid with n turns/m, length l, current i, and cross sectiona l A a magnetic field inside the solenoid, parallel to its axis with val ni B o This field creates a flux through the same device (the solenoid) that creates the field. Should the field change (maybe change the current), the flux will change and there will be an induced voltage and current. This process is called self induction.
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Transcript of Self Inductance Consider a solenoid with n turns/m, length l, current i, and cross sectional area A....
Self Inductance
Consider a solenoid with n turns/m, length l, current i, and cross sectional area A.
l
A
There is a magnetic field inside the solenoid, parallel to its axis with value:
niB o
This field creates a flux through the same device (the solenoid) that creates the field.Should the field change (maybe change the current), the flux will change and there will be an induced voltage and current.This process is called self induction.
What is the total flux, B ,through the solenoid?
AninlB 0
LiABNB ))()((
iL B
measured in Henrys
If the current changes, then Faraday’s law gives:
dt
diL
dt
d B
If L is not easily calculable from 1st principles, we can measure it using the above equation:
dtdi
L
Both quantities on rhs are easy to measure
The circuit symbol for an inductor is:
When the current is changing in the inductor, there will be a voltage drop across it:
dt
diLV
Compare this with the other circuit devices you have studied:
battery
iRV resistor
C
qV capacitor
RL CircuitR
ЄL
At t = 0 we close the switch up to put battery in series with L and R.What does Kirchhoff say?
i
0dt
diLiR
Energy in Inductors and Magnetic Fields
Let’s take the loop equation for the building inductor circuit and multiply by the current:
02 dt
diiLRii
Power supplied by battery
Power consumed by resistor
Power consumed by inductor
2
2
1Li
dt
d
dt
diiL
dt
dUP LL
Thus 2
2
1LiU L
For a solenoid: lAnL 20
niB o2
0
20
2 )(2
1
2
1
n
BlAnLiU L
AlB
0
2
2
Dividing by the volume Al, we get the energy density in the magnetic field:
0
2
2B
uB
Exercise; Suppose a large inductor of 10 H is effectively connected in series with a small resistance of of 1Ω. A knife-blade switch is used to turn the circuit on and off, and eventually a current of 10 A is established in the inductor. Why might the person turning the circuit off want to have a lot of insurance?
The energy stored in the inductor is:
JU L 50010102
1 2
As the switch is opened quickly, a huge voltage will develop. Why?Large di/dtThis energy will be discharged in about:
sec10
1t
