The coil opposes the change of current flowing through it. This property of the coil is called Self-inductance. The opposition offered by a coil depends on the rate of change of the current. The more the rate of change of current, the more the opposition. Why does this opposition to the current take place? The opposite polarity voltage is induced across the coil, and the magnitude of the voltage induced across the coil depends on the rate of change of current and the self-inductance of the coil. The induced EMF in a coil always opposes the applied voltage.
Self-inductance is defined as the induction of voltage in a current-carrying wire when the current in the wire itself changes.
The voltage induced in the current-carrying conductor is;
Thus, the voltage induced in the coil is directly proportional to the number of turns(N) and the rate of change of flux(dΦ/dt). The voltage induced in the coil opposes the applied voltage.
Expression for Self Inductance
We can find the self-inductance of the coil by the following expression.
If e= 1 Volt, dI/dt = 1 amp/sec,
Then L= 1 Henry
If the voltage induced in the coil is 1 volt when the current flowing through it changes at a rate of 1 Ampere/second, the coil has an inductance of 1 Henry.
The basic unit of inductance is Henry(H), which was named after Joseph Henry.
1 Henry = 1 Weber per Ampere
The self-inductance of a coil can also be expressed as;
The self-inductance of the coil can also be expressed as;
Where,
N = Number of turns
Φ = Flux
I = Current flowing through the coil
Self-Inductance Formula Derivation
Where,
μo is permeability of free space( 4π x 10-7 )
Factors on Self Inductance Depends
From the above expression, it is clear that the self-inductance of the coil depends on the;
- Permeability of the core
- Number of turns in a coil
- Cross-section area of the coil wire
- Length of the coil
Solved Problems on Self Inductance
In a coil, when the current changes from 5 A to 2 A in 0.1 s, the induced EMF is 3 Volts. What is the coil’s self-inductance?
E=-L(dI/dt)
E=-L[(I2-I1)/(t2-t1)]
3=-L[(2–5)/(0.1–0)]
3= -L(-3/0.1)
3 = 30 L
L= 3/30
L = 0.1 Henry
What is the self-inductance of a coil when a change of current from 0 to 1A in 0.05s induces an emf of 20 volts?
E=-L(dI/dt)
E=-L[(I2-I1)/(t2-t1)]
20=-L[(0–1)/(0.05–0)]
20= -L(-1/0.05)
20 = 20 L
L= 20/20
L = 1 Henry
An inductor coil has 400 turns of copper wire, which produce a magnetic flux of 20mWb when a DC current of 10 amperes flows through it. What is the coil’s self-inductance?
N = 400 Turns
I = 10 Amps.
Φ = 20mWb= 20 x 10-3
= 0.02 Wb
L = N Φ / I
L= 400 x 0.02/ 10
L= 0.8 Henry
