Inductance Calculator

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**L = $\frac{\mu N^2 A}{l}$**Where $\mu$ is the core material permeability,

**Step 1 :**

**Step 2 :**

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.

So the inductance is given by

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.

Read the given problem and observe the quantities given

Use the formula

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

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

A inductor coil has 600 turns of copper wire with core area 0.6 m

^{2}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 m^{2},

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.

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 = $\pi$ $\times$ r^{2}= 314.2 m^{2},

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.