Question

What is self-inductance? Establish expression for self-inductance of a long solenoid.

Answer 1

Hint: Self-inductance is the emf induced $e$ in a current carrying wire provided that the current $I$ is varying at a unity rate. In this we first start with the formula of magnetic flux density B that is $B = \dfrac{{{\mu _0}NI}}{l}$ then we substitute this in the equation of magnetic flux that is $\phi = BA$ and get$\phi = \dfrac{{{\mu _0}NI}}{l}A$. Now we equate $\phi = \dfrac{{{\mu _0}NI}}{l}A$ and $\phi = LI$. At last we get the required expression for self-inductance of a long solenoid as $L = \dfrac{{{\mu _0}NA}}{l}$.

Complete Step-by-Step solution:
Self-inductance is the emf induced $e$ or the voltage induces in a current-carrying wire provided that the current $I$ in the wire is varying at a unit rate that is $\dfrac{{dI}}{{dt}} = 1$. The expression for self-inductance $L$ is given as:
$L = \dfrac{e}{{\dfrac{{dI}}{{dt}}}}$------------------------------------- (1)
When $\dfrac{{dI}}{{dt}} = 1$ we get $L = e$. Its S.I unit is henry (H).
Now let us assume a long solenoid of length l, radius r, and the number of turns equal to N as shown in figure 1. Current $I$ flowing through the solenoid. The flux development is $\phi $ and the emf induced due to this varying current is $e$.

Figure 1

Now we know the formula for magnetic flux density is
$B = \dfrac{{{\mu _0}NI}}{l}$---------------------------------- (2)
Where B is magnetic flux density in tesla
             ${\mu _0}$ is the permeability of free space
We also know that the expression for magnetic flux is
$\phi = BA$------------------------------------ (3)
Where A is the cross-section area of the solenoid which is equal to $\pi {r^2}$. Now substituting equation number (2) in equation number (1) we will get
 $\phi = \dfrac{{{\mu _0}NI}}{l}A$---------------------------------- (4)
There is another expression for the flux magnetic that is:
$\phi = LI$-------------------------------------- (5)
Therefore, by equating equation (4) and equation (5) we will get
$LI = \dfrac{{{\mu _0}NI}}{l}A$
$ \Rightarrow L{I} = \dfrac{{{\mu _0}N{I}}}{l}A$
$ \Rightarrow L = \dfrac{{{\mu _0}NA}}{l}$
Hence the expression for self-inductance of a long solenoid is $L = \dfrac{{{\mu _0}NA}}{l}$

Note: For these types of questions we need to have a clear understanding of electric field intensity E, electric flux density D, Magnetic field intensity H, magnetic flux density B, and magnetic flux $\phi $. We need to know the expression for self-inductance and mutual-inductance when there is a solenoid, current-carrying wire, or a toroid.