- There is one coil in which there is change in the flux in that coil and because of that flux change.
- Current is induced in the same coil.
- Current tries to oppose the change in the flux.
- Consider a closed circuit,as a result the current will flow through the coil,therefore flux increases as a result current is induced in the coil.
- This induced current will oppose the growth of current.
- Suppose there is N number of turns in the coil.Therefore flux linkage of the coil N Φ∝ I.
- N Φ =LI where L=constant of proportionality and is known as self-inductance.
- Therefore Self-inductance will describe about the ratio of magnetic flux to the current it induces.
- Induced Emf e=-(d Φ/dt) By faraday’s Lenz’s law
- Therefore e=-d/dt [LI]
- e= -L dI/dt Where I=current flows through the coil.
- This Emf will oppose the change in I.
Self-inductance of a long solenoid
- Long solenoid is the one whose length(length) is very large as compared to radius(r) of the solenoid.(l>>r)
- Using B=μ0nI
- =(μ0NI)/length where N= total number of turns, n= number of turns per unit length.
- Case 1:- core of the solenoid has air.
- =>Φ =(μ0N2IA)/length (equation(1))
- Also total flux=LI (equation(2))
- From (1) and (2)
- => LI =(μ0N2IA)/length
- =>L = (μ0N2A)/length
- Case 2:- core of the solenoid is made of material which has permeability μr.
- L=(μr N2A)/length
- =>L= (μ0 μr N2 A)/length
Problem:- A long solenoid with 15 turns per cm has a small loop of area 2.0 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer: - Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A =2.0 cm2= 2 × 10-4m2
Current carried by the solenoid changes from 2 A to 4 A.
Therefore change in current in the solenoid, di = 4 – 2 = 2 A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as:
e= (dΦ/dt) … (i)
Where, Φ= Induced flux through the small loop
= BA … (ii) Where B = Magnetic field
= (μ0ni) … (iii)
μ0 = Permeability of free space
Hence, equation (i) reduces to:
e= (d/dt) (di/dt)
=Aμ0n x (di/dt)
=2x10-4x4x3.14x10-7x 1500 x (2/0.1)
=7.54 x 10-6V
Hence, the induced voltage in the loop is 7.54x10-6V