1. Two solenoids, A and B, are wound using equal lengths of the same kind of wire. The length of the axis of each solenoid is large compared with its diameter. The axial length of A is twice as large as that of B, and A has twice as many turns as B. What is the ratio of the inductance of solenoid A to that of solenoid B?
2. Homework Equations
L = μ0N2A/L
where N is the number of windings, A is cross-sectional area, and L is the axial length.
The Attempt at a Solution
I started by setting the inductance of solenoid B to LB = μ0N2A/L, and altering this equation for the dimensions of solenoid A as specified in the question such that
LA = μ0(2N)2A/2L = μ02N2A/L
in which case the ratio of A:B is 2.
However, I understand that because the question specifies that the same amount of wire is used for both solenoids, changing the length and winding number of solenoid A would also affect its cross-sectional area, but I'm not sure how it can be calculated.
If I calculated correctly, doubling the height of a cylinder but keeping volume constant would require the cross-sectional area to be decreased by half. In this case the inductance ratio of A:B would just be 1, but I don't think that's right.