The self-inductance of a coil (approximated as a solenoid wound on a core) is given by the formula:

$L = \mu n^2 A l$

Where:

$L$ is the self-inductance.
$\mu$ is the permeability of the core material.
$n$ is the number of turns per unit length.
$A$ is the cross-sectional area of the core.
$l$ is the length of the coil (which is the length of the core along the axis of the coil).

Let's analyze how each parameter changes based on the problem statement:

Linear dimensions of the core: The problem states that "all the linear dimensions of core are increased by a factor 3".

A core with a rectangular cross-section has three main linear dimensions: width ($w$), height ($h$), and length ($l$).

So, the new dimensions will be:
- New width $w' = 3w$
- New height $h' = 3h$
- New length $l' = 3l$
Cross-sectional area ($A$):

The original cross-sectional area was $A = w \times h$.

The new cross-sectional area $A'$ will be $A' = w' \times h' = (3w) \times (3h) = 9wh = 9A$.

So, $A'$ increases by a factor of 9.
Length of the coil ($l$):

As part of the linear dimensions of the core, the length of the coil (the length over which the turns are wound) also increases by a factor of 3.

So, $l' = 3l$.
Number of turns per unit length ($n$):

The problem explicitly states that "number of turns per unit length of coil remains same".

So, $n' = n$.
Permeability ($\mu$):

The core material is not changed, so its permeability $\mu$ remains constant.

Now, let's calculate the new self-inductance ($L'$):

The original self-inductance was $L = \mu n^2 A l$.

The new self-inductance is $L' = \mu (n')^2 A' l'$.

Substitute the changed parameters:

$L' = \mu (n)^2 (9A) (3l)$

$L' = \mu n^2 A l \times (9 \times 3)$

$L' = \mu n^2 A l \times 27$

Since $L = \mu n^2 A l$, we can write:

$L' = 27L$

Therefore, the self-inductance increases by a factor of 27.

The final answer is $\boxed{27}$.

Electromagnetic Induction

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