# Ratio of Inductance between 2 Solenoids?

• David Day
In summary, the ratio of the inductance of solenoid A to that of solenoid B is 2, taking into account the specified dimensions and the amount of wire used for each solenoid. However, the cross-sectional area of solenoid A may also be affected and cannot be accurately calculated without more information.

## Homework Statement

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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

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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.

David Day said:
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,
Yes.
but I'm not sure how it can be calculated.
Can you express the cross-sectional area in terms of the length of wire and the number of turns of wire?

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.
There is no requirement that the volumes of the cylinders be the same.

TSny said:
Yes.
Can you express the cross-sectional area in terms of the length of wire and the number of turns of wire?

There is no requirement that the volumes of the cylinders be the same.

Yeah, I was thinking that using the same amount of wire, the volume would be constant, which isn't actually the case.

So it seems to me that if the wire is of length x, and the circumference of the wire is 2πrN for each uniform winding, then x = 2πrN and r = x/2πN. I'm not sure if this is correct, though.

David Day said:
x = 2πrN and r = x/2πN.
Looks right.

• David Day