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Subsections

# 1 The Calculation

## 1.1 Introduction

We wish to calculate the magnetic field of a long solenoid with N turns per unit length carrying a current which has the same value at every point of the solenoid as in Jackson problem 6.24. We will solve for the case where the current is ReIe - it, and use this result to calculate the field the of triangular wave current, which is periodic and has the value.

## 1.2 The Green's function

The solution to the two-dimensional Helmholtz equation is

 + k 2 = - 4( - ) (1)

which has only outgoing waves at infinity is

 G(,,',') = iJm(k)Hm(1)(k)e im( - ') . (2)

## 1.3 The magnetic Field

For the solenoid the vector potential is then

 A = J1(k)H1(1)(k) (3)

where is the smaller of a and , and is the larger. Taking curl of , we get

 Bz = (A) . (4)

We can use the identify for Bessel functions C1 , that

 [xC1(x)] = C0(x) (5)

to give the magnetic field

 Bz = (6)

As a check, if 0, J1(ka) 0, and the > a component goes to zero, while H1(1)(ka) - 2i/(ka) , and we get the expected quasistatic result, Bz = 4NI()/c for < a .

## 1.4 Triangular Current Results

For a triangular wave current, I have
 I(t) = = Re (7)
where 2/ is the period of the oscillation. Plugging this in, the magnetic field as a function of time becomes

 Bz(,t) = Re e - int (8)

Defining x = /a , = a/c , = t , and B0 = 4NI0/c this can be written

 Bz(,t) = (9)

Next: 2 Low frequency results Up: The external magnetic field Previous: The external magnetic field