# Homework Help: Solenoids and Toroids

1. Jun 28, 2008

### Chris W

Hi everyone.
I am working on problems with solenoids and Toroids
I have solution for the solenoid:
B = μo i n

And toroid:
B = (μ o i n)/ (2Π r)
Also, I know that the magnetic field is the function of r namely: B = B(r)

r- radius of the Ampere’s path
n – number of loops per unit length
i-Current
μo – constant

My problem is:
Using the solution for the toroid, show that for the large toroid the answer can be approximated as the solenoid on the very small piece of the toroid.
I know that I have to play with limits. Something like:
a - inner radius of the toroid,
b – outer radius of the toroid,
I think I have to take a limit when ∆a goes to 0 and in this way radius a will approach radius b. in this way the solution for the toroid SHOULD be the solution for the solenoid (on the small length L of course)
I don’t know how to set it up. How to get from the toroid solution to the solenoid solution using limits or (other technique)

Thanks for help
Chris W

2. Jun 28, 2008

### nrqed

You should be more specific. The B fields you gave are the B fields evaluated where exactly? And your two equations can't both be right since they don't even have the same dimensions. There is something missing in your toroid expression.

(Hint: the two "n" do not have the same meaning)

3. Jun 28, 2008

### Chris W

Thanks nrqed

hmmm... I see it now... yeah...

-ok the calculated B field is in the toroid between the radius a and radius b.
-as for the meaning of n?? Yeah you are correct! there are two different n's

-I checked my math and I see now that the correct values for the B fields are:

for the Solenoid

B=μo*i*n

n - # of turn/ unit length
μo - constant (mu subscript o)

For the Toroid:

B=(μo*i*N)/(2*pi*r)

also, the strength of the B field id the function of r: B=B(r)

r - radius to the Ampere's Path a<r<b
N - # total number of loops in the toroid

__________________________________________________________________________
I was thinking about it and looks like I can say that:

a- inner radius of the toroid
b- outer radius of the toroid
(delta a) = b-a
r - radius of the Ampere's Path a<r<b
so...

b=(delta a)+a

limit when (delta a) goes to 0, than a=b
and I am stuck here...

___________________________________________________________________________
The goal is to use the solution of B field for the toroid ( INSIDE the toroid) and by taking the correct limit (or using other method) to obtain the solution for the B field of the solenoid INSIDE the SOLENOID.
So basically when (Delta a) is very small (limit goes to 0) over the small, call it; ds over solution looks like the regular, finite, straight solenoid with the B field inside.

Once again thanks a lot nrqed
Your comment opened my eyes a little bit.
I am still stuck, but I feel I am a step closer.

Any ideas? Need better explanation? I can derive the B fields for the solenoid and toroid if needed.

I can be more specific if needed.

Chris W
_____________________________________________________________________________

4. Jun 28, 2008

### nrqed

It's much simpler than this. In the limit r>>a and r >>b, you can basically say that the length of the toroid is 2 pi R, right? (That's the length of the path for Ampere's law but that's basically the length of the toroid as well in the limit r much larger than a and b).

Now, the answer falls off directly (paying attention to the difference between "n" appearing in the solenoid formula and the "N" appearing in the toroid formula).

5. Jun 28, 2008

### Chris W

WOW.... great

Thanks
This is easy!!

Thank you so much !!!!

you rock man!

Chris W

6. Jun 28, 2008

### nrqed

You are very welcome!

7. Jun 29, 2008

### Chris W

Hi all. Hi nrqed!
I Think I have a better way to do it!!!!!
hmmm... I was thinking about it and I think I see another way to do it.

the objective is to prove that the solution for the B field of toroid (with small delta a) is the solution of the solenoid (for the B field inside)

Also, I know that

N=n*A*L

so putting all together we have:

N= nAL n=N/AL

B = (μo i N)/(2 Π r) B = μo i n
B = (μo i n A L)/(2 Π r) B = (μo i N)/(A L)

B=B

o i n A L)/(2 Π r)=(μo i N)/(A L)

o i n A L)/(2 Π r)=(μo i N)/(A L)

( n A L)/(2 Π r)=( N)/(A L)

N= nAL
AL = 2 Π r
This is as far as I can go

But… if I say that AL=1 ( and I don't know WHY I can say that...lol)

( n )/(2 Π r)=N

N - # total number of loops in the toroid
n - (# of loops)/(unit length)

let’s check the units

N = (# of loops)/(unit length) * unit length= # of loops ≡ N

I hope I am doing this right.
Nrqed any input???

Thanks

Chris W

8. Jun 29, 2008

### Chris W

This section is of course:

N= nAL

n=N/AL

it came out together ....sorry
but N is defined above so there shouldn't be any problems
CHRIS W

9. Jun 29, 2008

### Chris W

Same here:

Toroid Solenoid
B = (μo i N)/(2 Π r) B = μo i n
B = (μo i n A L)/(2 Π r) B = (μo i N)/(A L)

10. Jun 29, 2008

### Chris W

Man... editing in this form is a challenge

Toroid
B = (μo i N)/(2 Π r)
B = μo i n

Solenoid
B = (μo i n A L)/(2 Π r)
B = (μo i N)/(A L)

11. Jun 29, 2008

### Chris W

I hope you can follow my math here ....
Thanks
Chris W

12. Jun 29, 2008

### nrqed

I don't follow. I don't even know what A is!

The only way to do the problem that I can see is what we discussed yesterday. The only key point is that you must define n= N/(length of the toroid) in order to compare with the solenoid result. But of ocurse, the length of a toroid is not well-defined.

Yesterday, I should NOT have written r>>a and r>>b, that was a mistake. Really, we are taking the limit r>>(b-a) (which also implies a>> (b-a) and b>>(b-a) . I apologize for this mistake.

For a toroid we can define the "inner length" 2pi a, or the "middle length" 2pi r or the outer length 2pi b. In the above limit, they are equal to one another,

For example, write r = a + (b-a)/2 Then clearly 2pi r approaches 2pi a in the limit above

$$2 \pi r = 2 \pi (a + \frac{a-b}{2}) \approx 2 \pi a$$

Patrick

13. Jun 29, 2008

### Chris W

Thanks Patric!

Yeah I noticed that something is wrong with r>>a and r>>b.
All good man!

Thanks!

yeah a good sport!

Chris W