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Inductance Calculator
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Its known that inductance (L) is the opposing change in current. Lets take coil wounded by the wire the resisting force that acts in a coil of wire due the current flow leads to the emf.
Inductance in a Coil
So the inductance is given by
L = $\frac{\mu N^2 A}{l}$Where $\mu$ is the core material permeability,
N is the no of turns the coil wire has,
A is the core area,
l is the average coil length.

Inductance Calculator is a online tool to calculate any unknown quantity of these inductance (L), core material permeability ($\mu$), no of turns (N), core area (A), average coil length (l) if any four of these are known.
 

Steps for Inductance Calculator

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Step 1 :  

Read the given problem and observe the quantities given



Step 2 :  

Use the formula


L = $\frac{N^2 \mu A}{l}$


Substitute the given values in above equation and calculate the unknown quantity.



Problems on Inductance Calculator

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  1. A inductor coil has 600 turns of copper wire with core area 0.6 m2 and length 70 m with permeability 0.6.Calculate the self-inductance.


    Step 1 :  

    Given: No of turns N = 600 turns,
    Area A = 0.6 m2,
    Length l = 70 m,
    Permeability $\mu$ = 0.6



    Step 2 :  

    Self Inductance is given by
    L = $\frac{\mu N^2 A}{l}$
      = $\frac{0.6 \times 600^2 \times 0.6 m}{70 m}$
      = 1851.4 Henry



    Answer  :  

    Hence the inductance L of the coil is 1851.4 Henry.



  2. A coiled wire has 320 turns having the diameter of 20 cm and an unstretched length of 29 cm. what is its self inductance?


    Step 1 :  

    Given: No of turns N = 320 turns,
    Area A = $\pi$ $\times$ r2 = 314.2 m2,
    Length l = 70 m,
    Permeability $\mu$ = 0.6



    Step 2 :  

    Self Inductance is given by
    L = $\frac{\mu N^2 A}{l}$
      = $\frac{0.6 \times 320^2 \times  314.2 m}{70 m}$
      = 275777.8 Henry



    Answer  :  

    Hence the inductance L of the coil is 275777.8 Henry.



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