Inductor Loop Calculations

1. Jun 22, 2014

Tom_Greening

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

1. The problem statement, all variables and given/known data
A single loop carries 100mA and has an inductance of 100μH

2. Relevant 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?

3. 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

Last edited: Jun 22, 2014
2. Jun 23, 2014

Staff: Mentor

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.

3. Jun 23, 2014

Tom_Greening

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?

4. Jun 23, 2014

Staff: Mentor

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.

5. Jun 23, 2014

Tom_Greening

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)?

6. Jun 23, 2014

Staff: Mentor

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.

7. Jun 23, 2014

Tom_Greening

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.

8. Jun 23, 2014

Staff: Mentor

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

9. Jun 23, 2014

Tom_Greening

Ah, got it. Thanks so much for the assistance NascentOxygen