# How Does Coil Configuration Affect Solenoid Inductance?

• Rayanna
In summary, we have to assume that the two coils are made of copper wires of the same length, and that they are coiled with an equal number of turns per unit length. This allows us to compare the inductances of the two coils, which are given by the formula μN^2A/l, where A is the area of the cross section of the solenoid frame. By putting in the given values, we can see that the ratio of self inductances is 3:1, which is different from the 1:1 ratio that we initially get if we assume the solenoids have the same length. This is because coiling the wires equally will result in different lengths of the solenoids, leading
Rayanna
Two coils are made of copper wires of same length .In the first coil number of turns is 3n and radius is r . In the second coil number of turns is n and radius is 3r the ratio of self inductances of the coil is:

I know that self inductance of a solenoid is μN2A/l ;
where A = area of cross section of frame of solenoid ;so by puting values I get 1:1 which is incorrect ? Please explain me where I am wrong as the correct answer is 3:1.

Last edited:
Rayanna said:
Two coils are made of copper wires of same length .In the first coil number of turns is 3n and radius is r . In the second coil number of turns is n and radius is 3r the ratio of self inductances of the coil is:

I know that self inductance of a solenoid is μN2A/l ;
where A = area of cross section of frame of solenoid ;so by puting values I get 1:1 which is incorrect ? Please explain me where I am wrong.

It's a sort of trick question. Hint: what is "length" in this context?

PeroK said:
It's a sort of trick question. Hint: what is "length" in thi
PeroK said:
It's a sort of trick question. Hint: what is "length" in this context?
I am not sure but may be copper wires are of same length.
But then I can't find the length of solenoid

##l## is the length of the solenoid. Is that the same in both cases? The total length of the wire is the same.

Also, what is ##N##? Is that the same in both cases?

PeroK said:
##l## is the length of the solenoid. Is that the same in both cases? The total length of the wire is the same.

Also, what is ##N##? Is that the same in both cases?
I am also not sure that length of solenoid is same, as it is not cleary stated in question that whether copper wires used to make coil are of same length or coils are of same length but N in first case is 3n and in second case is n.

Rayanna said:
I am also not sure that length of solenoid is same, as it is not cleary stated in question that whether copper wires used to make coil are of same length or coils are of same length but N in first case is 3n and in second case is n.

You are told that the wire is the same length in both cases, and I think you are supposed to assume that the wire is coiled with the same number of coils per unit length in both cases. This would give you:

##r_1 = r, \ N_1 = 3n, \ l_1 = l##

##r_2 = 3r, \ N_2 = n, \ l_2 = l/3##

The solenoid must be shorter in th second case, where the length is each coil is longer.

PeroK said:
You are told that the wire is the same length in both cases, and I think you are supposed to assume that the wire is coiled with the same number of coils per unit length in both cases. This would give you:

##r_1 = r, \ N_1 = 3n, \ l_1 = l##

##r_2 = 3r, \ N_2 = n, \ l_2 = l/3##

The solenoid must be shorter in th second case, where the length is each coil is longer.
Thanks for the solution.
But still I did not understand the point that why we should assume that number of turns per united length is same?

Rayanna said:
Thanks for the solution.
But still I did not understand the point that why we should assume that number of turns per united length is same?

If you make the solenoid the same length in both cases, then your 1-1 ratio is correct. It might seem quite natural to do this in order to comapre the inductance in the two cases.

But, another natural approach is to coil the wire equally in both cases (probably tightly). This leads to different lengths of solenoid in the two cases.

On reflection, I would say the second assumption is better, because you can definitely do that. In the first case, you may have to stretch the wire and the loops are less and less like circles if the radius is large.

A diagram might help.

Rayanna
PeroK said:
If you make the solenoid the same length in both cases, then your 1-1 ratio is correct. It might seem quite natural to do this in order to comapre the inductance in the two cases.

But, another natural approach is to coil the wire equally in both cases (probably tightly). This leads to different lengths of solenoid in the two cases.

On reflection, I would say the second assumption is better, because you can definitely do that. In the first case, you may have to stretch the wire and the loops are less and less like circles if the radius is large.

A diagram might help.
Thank you.

I get it so we have to assume both copper wires are of same length and we have coiled them in such manner that there number of turns per unit length are same.

## What is self induction?

Self induction is the phenomenon in which a changing magnetic field within a closed circuit induces a voltage within that same circuit.

## How does a solenoid exhibit self induction?

A solenoid is a coil of wire that produces a strong magnetic field when an electric current is passed through it. When the current changes, the magnetic field also changes, inducing a voltage within the solenoid itself.

## What factors affect the amount of self induction in a solenoid?

The amount of self induction in a solenoid depends on the number of turns of wire, the strength of the magnetic field, and the rate of change of the current.

## What is the formula for self inductance in a solenoid?

The formula for self inductance in a solenoid is L = μN²A/l, where L is the self inductance, μ is the permeability of the core material, N is the number of turns of wire, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.

## How is self induction used in practical applications?

Self induction is used in a variety of applications, including inductors in electronic circuits, transformers, and generators. It is also used in devices such as relays and motors to control the flow of electricity and create motion.

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