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 this opposition of 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.The Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing.

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**

The self inductance of the coil can be found by following expression.

If e= 1 Volt, dI/dt = 1 amp/sec, then L= 1 Henry

The coil has inductance of 1 Henry, if voltage induced in the coil is 1 volt when the current flowing through the coil changes at a rate of 1 Ampere/second.

The basic unit of inductance is Henry(H) after Joseph Henry.

** 1 Henry = 1 Weber per Ampere**

**Solved problems of self Inductance**

**In a coil, when current current changes from 5 A to 2 A in time 0.1 s,induced EMF is 3 Volts. What is the self inductance of the coil?**

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

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

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 DC current 10 amperes flows through the coil. What is the self inductance of the coil?**

N = 400 Turns

I = 10 Amps.

Φ = 20mWb= 20 x 10

Φ = 20mWb= 20 x 10

^{-3}^{ }^{= 0.02 Wb}L = N

**Φ**/ IL= 400 x 0.02/ 10

**L= 0.8 Henry**

**The self inductance of a coil can also be expressed as;**

The self inductance of the coil can be expressed with above expressions.

Where,

N = Number of turns

Φ = Flux

I = Current flowing through the coil

**Factors on Self Inductance Depends **

Where,

μo is permeability of free space( 4π x 10

^{-7 })From above expression it is clear that the self-inductance of the coil depends on;

**1. Permeability of the core**

2. Number of turns in a coil

3. Cross section area of the coil wire

4. Length of the coil

2. Number of turns in a coil

3. Cross section area of the coil wire

4. Length of the coil

**Related Posts**

**Resistance and Leakage Reactance or Impedance of Transformer****What is Magnetic reluctance – Definition**

Please follow and like us: