Consider an inductor whose linear dimensions are tripled and the total number of turns per unit length is kept constant, what happens to the self-inductance?

The correct option is 3 times.

To understand why this is the case, we look at the formula for the self-inductance ($L$) of an inductor (such as a solenoid):

$L=\frac{{\mu }_0{N}^{2}A}{l}$

where:
- ${\mu }_0$ is the permeability of free space,
- $N$ is the total number of turns,
- $A$ is the cross-sectional area of the inductor ($A=\pi r^2$, where $r$ is the radius),
- $l$ is the length of the inductor.

When the linear dimensions of the inductor are tripled, both its radius and length are scaled by a factor of 3:

$r^{\prime }=3r$

$l^{\prime }=3l$

This affects the cross-sectional area ($A^{\prime }$) as follows:

$A^{\prime }=\pi {\left(r^{\prime }\right)}^2=\pi {\left(3r\right)}^2=9\pi r^2=9A$

Since the total number of turns $N$ remains constant, we can substitute the new dimensions into the inductance formula to find the new self-inductance ($L^{\prime }$):

$L^{\prime }=\frac{{\mu }_0{N}^2{A}^{\prime }}{l^{\prime }}=\frac{{\mu }_0{N}^2\left(9A\right)}{3l}=3\left(\frac{{\mu }_0{N}^{2}A}{l}\right)=3L$

Therefore, the self-inductance becomes 3 times its original value.