Question

derive expression for solenoid and toroaid.

Answer 1

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)

Answer 2

Answer:

Let r be the average radius of the toroid and n be the number of turns per unit length and N = 2πrn = (average) perimeter of the toroid × number of turns per unit length. On comparing the two results: for a solenoid and toroid. Equations (1) and (3) will give, we get B = μ0 n I, i.e., the result for the solenoid.