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

`(L_1)/(L_2) =(1)/(3)`

Step-by-step explanation:

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

`L=(\mu_0\,N^2\,A)/(I)`

Then, for the first coil we have;

`L_1=(\mu_0\,N^2\,A)/(I)`

and for coil 2 we have:

`L_2=(\mu_0\,(6\,N)^2\,(A/4))/(3\,I)`

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

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