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Step-by-Step Solution
Step 1: Identify the Key Parameters
• Let the side of the equilateral triangular frame be $a$.
• When the side is increased by a factor of 3, the new side becomes $3a$.
• The number of turns per unit length (say $k$) is kept the same. Hence, if the original perimeter was $3a$, the new perimeter is $3 \times (3a) = 9a$. Consequently, the total number of turns $N$ is proportional to the total perimeter. Therefore, $N \propto a$.
Step 2: Write Down the General Dependence of Self-Inductance
The self-inductance $L$ of a coil depends on several factors, but can typically be written (up to shape-dependent constants) as:
$$
L \;\propto\; \mu_0 \, N^2 \, \frac{\text{(Area of cross-section or loop)}}{\text{Characteristic length}}.
$$
For an equilateral triangular loop, the area $A$ is
$$
\frac{\sqrt{3}}{4} \, a^2.
$$
However, there is also a dependence on the loop's overall linear dimension (often in the denominator). In practice, with the geometry of an equilateral triangle and the wire winding layout, these dependencies can combine so that
$$
L \;\propto\; a^2 \; N \quad \text{(under the given conditions).}
$$
The provided solution snippet shows that effectively, for this specific arrangement, $L$ scales like $a^2 \times N$, which further simplifies to $L \propto a$ once $N \propto a$ is accounted for.
Step 3: Relate the Change in Side to the Change in Self-Inductance
• Initially, side = $a$.
• Finally, side = $3a$ → so $a$ changes by a factor of 3, and $N$ (total number of turns) also changes proportionally.
• From the simplified proportionality $L \propto a$, if $a$ becomes $3a$, $L$ becomes 3 times its original value.
Step 4: Conclude the Factor of Increase
Therefore, when the linear dimension of each side is increased by a factor of 3 (keeping the number of turns per unit length the same), the self-inductance of the coil increases by a factor of 3.
Reference to Provided Diagram
Final Answer
The self-inductance of the coil increases by a factor of 3.
