Question

While keeping the area of cross-section of a solenoid the same, the number of turns and length of solenoid one both doubled. The self-inductance of the coil will be

Answer 1

Hint : Use formula of self-induction. It depends on a number of turns, magnetic flux and current carrying in the conductor. Use the formula of magnetic induction in solenoid. Magnetic field depends on the number of turns per unit length and current carrying in solenoid.

Complete step by step solution :
The phenomena of the production of induced e.m.f. in a coil due to changes of current in the same coil, is called self-induction.
Mathematically, self-induction of a coil is given as,
$L=N\dfrac{\phi }{\dfrac{di}{dt}}=N\dfrac{\phi }{I}$
Where,
$\begin{aligned} & N=\text{number of turns on coil} \\\ & \text{A= cross sectional area} \\\ & \text{I= current passes through coil} \\\ \end{aligned}$
Now Magnetic flux associated with the coil is given as,
$\begin{aligned} & \phi =BA--(1) \\\ & \text{where,} \\\ & B=\text{magnetic induction} \\\ \end{aligned}$
We know that using ampere’s law, magnetic induction at a point well inside along the axis of solenoid is given as,
$\begin{aligned} & B={{\mu }_{0}}nI \\\ & \text{where,} \\\ & \text{I= current flowing through solenoid} \\\ \end{aligned}$
Put above value in equation one, we get
$\phi =({{\mu }_{0}}nI)A$
We know that $n$ is defined as the number of coils per unit length.
Mathematically,
$\begin{aligned} & n=\dfrac{N}{l} \\\ & \text{where,} \\\ & l=\text{length of solenoid} \\\ & \text{N= number of turns} \\\ \end{aligned}$
Finally self-induction is given as,
$L={{\mu }_{0}}\dfrac{{{N}^{2}}A}{l}$
Given that, number of turns and length of solenoid is double, i.e.
$\begin{aligned} & {{L}^{'}}={{\mu }_{0}}\dfrac{{{(2N)}^{2}}A}{2l}=2{{\mu }_{0}}\dfrac{{{N}^{2}}A}{l}=2L \\\ & \text{where,} \\\ & {{\text{L}}^{'}}=\text{ self induction when N and length}\left( l \right)\text{ gets double} \\\ \end{aligned}$
The self-inductance of the coil will be $2L$

Note : Self-induction is also called as back e.m.f. magnetic induction at a point well inside along the axis of solenoid is given as,$B={{\mu }_{0}}nI$. But magnetic induction at a point near the end of solenoid is given by$B=\dfrac{{{\mu }_{0}}nI}{2}$. According to change in magnetic induction, the value of self-induction will also change. Here. Do not get confused between$n\text{ and N}$. $n$is a number of turns per unit length. Whereas $N$is a number of turns per unit length.
Solenoid is cylindrical, like a structure having a number of turns on it.