Question

a current of 1.5 ampers in a 800 turns coil produces 1.5 × 10 to the power -5 weber magnetized flux in each other. Find the self - induction of coil.​

Answer 1

Given:

I = 1.5 A

N = 800

φ = 1.5 $\times 10^{-5$ Wb

To find:

L=?

Explanation:

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

In the case of self-inductance, the magnetic field created by a dynamic current in the circuit itself induces a voltage in the same circuit.

It is also defined as the magnetic flux passed through the coil due to current within the coil itself.

It is calculated by the formula

Nφ = LI

where

N = no of turns in the coil

φ - electric flux

L = self-induction of the coil

I = current passing through the coil

Put given values in above eq

800( 1.5 $\times 10^{-5$ ) = L ( 1.5 )

L = 8 $\times 10^{-3$

L = 8 mH

∴ The self-induction of the coil of the given coil is 8 mH.