Question

What will be the self-inductance of a coil of 100 turns if a current of 5A produces a magnetic flux of $5\times 10^{-5}$ Wb?
$\text{A.}\quad 1 mH$
$\text{B.}\quad 10 mH$
$\text{C.}\quad 1 \mu H$
$\text{D.}\quad 10 \mu H$

Answer 1

Hint: Self-inductance or the inductance of a coil is the property of the coil due to which it opposes the change of current flowing through it. Inductance is attained by a coil due to the self-induced emf produced in the coil itself by changing the current flowing through it.

Formula used:
$Total\ flux = Li$

Complete step by step answer:
According to Faraday’s law, if there is a change in flux in a coil, it will produce or generate an E.M.F. in the coil. According to Lenz law, the change in magnetic field is opposed in such a manner that the current inside the coil will resist the magnetic field by flowing in a particular direction. This property of coil is called self-inductance. Now we will calculate the self-inductance of a coil using: $Total\ flux = Li$
Now, flux produced = $5\times 10^{-5} Wb$
Hence total flux associated with 100 turns = $100\times 5\times 10^{-5} Wb = 5 \times 10^{-3} Wb$
Also, $i = 5A$
Now, putting values in $Total\ flux = Li$, we get;
$5\times 10^{-3} = L \times 5$
$\implies L = 10^{-3} H= 1mH$


Additional information:
Lenz law is just an extension of Faraday’s law. According to Faraday’s law, the E.M.F. produced inside a coil is has a magnitude equal to the rate of change of magnetic flux i.e. $|\varepsilon|= \dfrac{d\phi}{dt}$. But the direction in which the current will flow (i.e. clockwise or anti-clockwise) is given by Lenz law. Hence the complete law is $\varepsilon = -\dfrac{d\phi}{dt}$.

Note:
Magnetic flux, E.M.F and inductance are scalar quantities. These play a very important role in understanding the effects of magnetic flux or field and hence the generation of E.M.F. Also the magnetic field is a vector quantity. The S.I. unit of self-inductance(L) is Henry (H) whereas S.I. unit of magnetic flux = Weber (Wb).