Question

If all linear dimensions of an inductor are tripled, then self-inductance will become (keeping the total number of turns per unit length constant).

• A. $3times$
• B. $9times$
• C. $27times$
• D. $\frac{1}{3}times$

Answer 1

The correct option is C

$27times$

Step 1: Given data

All linear dimensions of an inductor are tripled.

To find the value of self-inductance.

Step 2: Formula used

Magnetic flux,$\varphi =\frac{\left({\mu }_{0}NiA\right)\times N}{l}$

where $N$ is the number of turns of the inductor, $i$ is current ${\mu }_0$ is the permittivity of free space, $l$ is length and $A$ is area of cross section.

Step 3: Calculate the Inductance

We know, $\varphi =LI$

where $L$ is inductance and $i$ is current

Therefore,

$$\begin{gathered} L=\frac{\varphi }{i} \\ L=\frac{{\mu }_0{N}^{2}A}{l} \end{gathered}$$

Number of turns per unit length, $n=\frac{N}{l}$

Area, $A=l\times l$

$$\begin{gathered} L=\frac{{\mu }_0{N}^{2}A}{l} \\ \Rightarrow L=\frac{{\mu }_0{\left(n\times l\right)}^2\left(l\times l\right)}{l} \\ \Rightarrow L={\mu }_0{n}^2{l}^3 \\ \Rightarrow L\propto l^3 \end{gathered}$$

Therefore, when the dimensions are tripled, self-inductance

$$\begin{gathered} L\text{'}\propto {\left(3l\right)}^3 \\ L\text{'}\propto 27l^3 \end{gathered}$$

Therefore, when the dimensions are tripled, self-inductance becomes $27times$.

Hence, option C is correct.