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Mashutka [201]
3 years ago
7

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 runni

ng through it, and coil 2 has current 3i running through it, what is ratio of their self inductances L1/L2
Physics
1 answer:
solmaris [256]3 years ago
7 0

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}

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