Question

Circular coil 1 has N turns and circular coil 2 has 6N. Coil 1 has area A and coil 2 has area A/4. If Coil 1 has current i running through it, and coil 2 has current 3i running through it, what is ratio of their self inductances L1/L2

Answer 1

Answer:

$\frac{L_1}{L_2} =\frac{1}{3}$

Explanation:

Recall that the formula for an inductance (L) for coil on N turns, are A and current I is given by:

$L=\frac{\mu_0\,N^2\,A}{I}$

Then, for the first coil we have;

$L_1=\frac{\mu_0\,N^2\,A}{I}$

and for coil 2 we have:

$L_2=\frac{\mu_0\,(6\,N)^2\,(A/4)}{3\,I}$

then, the quotient L1/L2 can be written as:

$\frac{L_1}{L_2} =\frac{\mu_0\,N^2\,A\,3\,I}{\mu_0\,(6\,N)^2\,(A/4)\,I}=\frac{12}{36} =\frac{1}{3}$