Question

A current of 1.5 amperes in a 400 turns coil
produces 0.4*10- weber magnetized flux in each turn. Find out the
self-inductance of coil.
​

Answer 1

Answer:

$\boxed{ \bf{n\:\phi = L.i}}$

Where,

'n' is the number of turns,

Ф refers to the magnetic flux,

L refers to the self inductance of the material,

'i' refers to the current flowing through the material.

Since we are required to find the value of 'L' we transpose all the other terms to the RHS. Hence we get:

$\implies L = \dfrac{n\:\phi}{i}\\\\\\\text{Substituting the values we get:}\\\\\\\implies L = \dfrac{400 \times 0.4 \times 10^{-4}}{1.5}\\\\\\\implies L = 106.67 \times 10^{-4}\\\\\\\implies \boxed{ \bf{L = 10.67 \:\:\textbf{mH}}}$

This is the required answer.

Answer 2

Given :-

A current t of 1.5 amperes in a 400 turns coil  produces 0.4 × 10- weber magnetized flux in each turn

To Find :-

Self inductance

Solution :-

We know that

$\bf n\theta = Li$

$\sf 400 \times 0.4 \times 10^{-4} = L(1.5)$

$\sf 160 \times 10^{-4} = 1.5L$

$\sf L = \dfrac{160\times 10^{-4}}{1.5}$

$\sf L = 10.7 \; mH$