Total length L will remain constant
\(L = (3a) N\) (N = total\, turns)
and length of winding (\(l\))= \((d) N\) (d = diameter of wire)
Self inductance = \(\mu_o n^2 Al= \mu_o n^2(\frac{\sqrt3 }4 a^2)dN\)
Self inductance (L) ∝ \(a^2N\)
As \(N = \frac{l}{3a} \rightarrow N ∝\frac1{a}\)
\(\therefore L ∝a\)
So, as a is increased to 3a the Self inductance will be 3 times.
Hence, the correct answer is option (B): Increases by a factor of 3
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where