Class 12 Physics Electromagnetic Induction | Self-Inductance |

**Self-Inductance**

- 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(l
_{ength}) is very large as compared to radius(r) of the solenoid.(l>>r) - Using B=μ
_{0}nI - =(μ
_{0}NI)/l_{ength}where N= total number of turns, n= number of turns per unit length. - Case 1:- core of the solenoid has air.
- Flux=NBA

- =NA((μ
_{0}NI)/l_{ength}). - =>Φ =(μ
_{0}N^{2}IA)/l_{ength}(equation(1)) - Also total flux=LI (equation(2))
- From (1) and (2)
- => LI =(μ
_{0}N^{2}IA)/l_{ength} - =>
**L = (μ**_{0}N^{2}A)/l_{ength} - Case 2:- core of the solenoid is made of material which has permeability μ
_{r}.- L=(μ
_{r}N^{2}A)/l_{ength} - =>
**L= (μ**_{0 }μ_{r }N^{2 }A)/l_{ength}

- L=(μ

** 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 cm^{2}= 2 × 10^{-4}m^{2}

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

= (μ_{0}ni) … (iii)

μ_{0} = Permeability of free space

=4nx10^{-7}H/m

Hence, equation (i) reduces to:

e= (d/dt) (di/dt)

=Aμ_{0}n x (di/dt)

=2x10^{-4}x4x3.14x10^{-7}x 1500 x (2/0.1)

=7.54 x 10^{-6}V

Hence, the induced voltage in the loop is 7.54x10^{-6}V

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