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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

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mberisso

Answer:

[tex]\frac{L_1}{L_2} =\frac{1}{3}[/tex]

Explanation:

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

[tex]L=\frac{\mu_0\,N^2\,A}{I}[/tex]

Then, for the first coil we have;

[tex]L_1=\frac{\mu_0\,N^2\,A}{I}[/tex]

and for coil 2 we have:

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

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

[tex]\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}[/tex]