Question

Q1.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

Step-by-step explanation:

Answer:

$\boxed{ \bf{n\:\phi = L.i}} </p><p>$

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:

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

This is the required answer.