What Happens to Inductance with Changes in Current and Loop Configuration?

AI Thread Summary
Inductance is independent of current, as indicated by the formula L = μ0 * (N^2*A/length), where L depends on the number of turns (N), area (A), and length. Increasing the current does not change the inductance but affects the magnetic flux through the loop. For one loop at 100mA, the inductance remains 100μH, while the flux changes with current. When the number of loops increases, the inductance increases proportionally, but the actual values depend on the specific area and length used in calculations. Understanding the relationship between these variables is crucial for accurate analysis in electromagnetism.
Tom_Greening
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Hello, I'm new to this forum. I hope I am submitting this to the right thread.
Just started physics and I am looking to get a little assistance

Homework Statement


A single loop carries 100mA and has an inductance of 100μH

Homework Equations


For air cored

L = μ0 * (N^2*A/length)

L = Nø/i

What is the inductance in each case?
a)What happens to inductance when the current is increased to 200mA? And what is the flux through loop?
b)What happens to inductance when the current is increased to 400mA and there are two loops, what is the flux through loop?
c)What happens to inductance when the current is set to 100mA and four loops, what is the flux through loop?

The Attempt at a Solution



There is 100mA and one loop = 100μH
Then
L = Nø / i
100μH = 1 turn*ø / 0.1A, therefore flux is equal to = 1x10-6

a) If I increase the current to 200mA then the inductance will be half based on formula.

1 turn*ø / 0.2A = 50μH, the flux remains the same = 1x10-6

b) If I increase the current to 400mA and there are two loops then I've effectively doubled the length, Area remains the same, but how do I calculate to new inductance and flux?c) If I leave the current at the original 100mA and put four loops, then I would multiply the inductance by 4 to get 400μH

I'm new to this and i am not sure if I am getting it right.

Many thanks
Tom
 
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Hi, and welcome to the Physics Forums.

L = μ0 * (N^2*A/length)

Is that N2 In that formula? So N is the number of turns ...

You can see there is no mention of current in that formula, meaning that L is independent of current.
 
NascentOxygen said:
Hi, and welcome to the Physics Forums.



Is that N2 In that formula? So N is the number of turns ...

You can see there is no mention of current in that formula, meaning that L is independent of current.

Ah ok! Yes, that is supposed to be N squared. I don't know how to enter the superscript. Thank you for the nudge in the right direction!

So let me do this again...

In the case of a)
The inductance will remain the same at 100μH, due the fact it is still the same area and same loop.
But the flux will be L = Nø/i => 100μH = 1* ø/200mA so flux = 20*10-6


In the case of b)
There are two loops, so I use L = μ0 * (N^2*A/length) where I'll let Area = 1 , length = 1 (But length is equal to 2 because 2 loops. So L = μ0 * (2^2*1/2) = L = 2.51μH
Then I put the inductance into L = Nø/i so 2.51μH=2*ø/0.4A so flux 0.50*10-6

In the case of c)
There are four loops, so L = μ0 * (4^2*1/4) so L = 5.02μH
Then inductance into L = Nø/i so 5.02μH=4*ø/0.1A so flux 0.125*10-6

I'm not sure if that is right though! Isn't the inductance meant to go up as turns increase?
 
Tom_Greening said:
In the case of a)
The inductance will remain the same at 100μH, due the fact it is still the same area and same loop.
Yes.

You don't know the absolute dimensions, but you could work out the value of the ratio area/length here. It might be useful later.
 
NascentOxygen said:
Yes.

You don't know the absolute dimensions, but you could work out the value of the ratio area/length here. It might be useful later.

I'm not quite sure I get it. I understand that if I let area A = 1 and say length = 1 for 1 loop then for 2 loops the A = 1 and length = 2 isn't it? So ratio would be 1/2? For four loops ratio is 1/4?

Isn't that what I did in case b)?
 
Tom_Greening said:
if I let area A = 1 and say length = 1
You won't be able to compare your "results" with their single-turn inductor because they didn't assume those unity values. Presumably they used the actual values. So you should, too, i.e., stay with their value of area/length ratio.
 
We weren't provided with area or length values. The question is as the lab cover sheet describes.

I suspect i could be misunderstanding the formulas...
The only other way to do algebraic ratios is define area/length as=> pi*r^2 / pi*2*r, but I don't think that works either.
I'm stumped.
 
You were given this equation:
L = μ0 * (N^2*A/length)

We can see it has 3 variables, if we regard area/length as 1 unknown here.

You are given a case where 2 of these are known, so you should be able to rearrange the equation to calculate that third one.

The formula itself is explained here: http://www.allaboutcircuits.com/vol_1/chpt_15/3.html
 
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Ah, got it. Thanks so much for the assistance NascentOxygen
 
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