Question

A wire of fixed length is wound on a solenoid of length $l$ and radius $R$. Its Self Inductance is found to be $L$. Now, if the same wire is wound on a solenoid of length $\frac{I}{2}$ and radius $\frac{R}{2}$, then self inductance will be

Answer 1

Step 1: Given

$\text{Original length}=l$

$\text{New length}=\frac{l}{2}$

$\text{original inductance}=L$

Step 2: To find

We have to find the new inductance.

Step 3: Calculate the new inductance

Use the formula for self-inductance.

The expression for self-inductance is given by,
$L=\frac{{\mu }_0{N}^{2}A}{l}...\left(1\right)$
Where, ${\mu }_0$ is the magnetic permeability of free space, $N^2$ is the number of turns per unit length, $A$ is the area of the solenoid, and, $l$ is the length of the solenoid.
The area of a solenoid is given by,

$A=\mathrm{\pi }{\mathrm{r}}^2$
Where, $\mathrm{r}$ is the radius of the cross-section of the solenoid.

Substituting this value in equation (1).

$L=\frac{{\mu }_0{N}^2\mathrm{\pi }{\mathrm{r}}^2}{l}...\left(2\right)$

Now, the length of the wire is given by,
$\mathrm{length}=\mathrm{circumference}\times \mathrm{number}\mathrm{of}\mathrm{turns}$
$\Rightarrow l=2\mathrm{\pi r}\times N$

Since the same wire is taken in both cases, then the circumference and number of turns will be the same for both wires.
$\Rightarrow L=2\mathrm{\pi r}\times N$

Rearranging the above equation.
$r=\frac{L}{2\mathrm{\pi }N}...\left(3\right)$
Now, substitute equation (3) in equation (2).

$L=\frac{{\mu }_0{N}^2\mathrm{\pi }{\left({\displaystyle \frac{\mathrm{L}}{2\mathrm{\pi N}}}\right)}^2}{l}$
$\Rightarrow L={\displaystyle \frac{{\mu }_0{L}^2}{4\pi l}}$
This will be the value of $L$ when the length of the wire is $l$ and radius is $\mathrm{r}$.
For, length ${\displaystyle \frac{l}{2}}$ and radius ${\displaystyle \frac{r}{2}}$ value of $L$ will be,
$L_0={\displaystyle \frac{{\mu }_0{L}^2}{4\pi {\displaystyle \frac{l}{2}}}}$

$\Rightarrow L_0=2{\displaystyle \frac{{\mu }_0{L}^2}{4\pi l}}$
Thus, the self-inductance of the coil will be doubled.

Hence, the new self-inductance will be two times the original self-inductance.