Solving Solenoids & Toroids: Show Approximation

[Chris W]

Hi everyone.
I need help please.
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,
∆a - the difference between radius a and radius b.
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

 

[nrqed]

Hi everyone.
I need help please.
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,
∆a - the difference between radius a and radius b.
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

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)

 

[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=μ_o*i*n

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


For the Toroid:

B=(μ_o*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.

Please help!
I can be more specific if needed.

Chris W
_____________________________________________________________________________

 

[nrqed]

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=μ_o*i*n

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


For the Toroid:

B=(μ_o*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.

Please help!
I can be more specific if needed.

Chris W
_____________________________________________________________________________

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).

 

[Chris W]

WOW... great

Thanks
This is easy!

Thank you so much !

you rock man!

Chris W

 

[nrqed]

WOW... great

Thanks
This is easy!

Thank you so much !

you rock man!

Chris W

You are very welcome!

 

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

B=B

(μ_o i n A L)/(2 Π r)=(μ_o i N)/(A L)

(μ_o i n A L)/(2 Π r)=(μ_o 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.
Can someone please check it?
Nrqed any input?

Thanks

Chris W

 

[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

 

[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)

 

[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)

 

[Chris W]

I hope you can follow my math here ...
anyone? please... is this correct?
Thanks
Chris W

 

[nrqed]

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

....

Thanks

Chris W

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

 

[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