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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 - i$\scriptstyle\omega$t, 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

$\displaystyle\nabla^{2}_{}$ + k 2 = - 4$\displaystyle\pi$$\displaystyle\delta$($\displaystyle\vec{r}$ - $\displaystyle\vec{r'}$) (1)

which has only outgoing waves at infinity is

G($\displaystyle\rho$,$\displaystyle\phi$,$\displaystyle\rho$',$\displaystyle\phi$') = i$\displaystyle\pi$$\displaystyle\sum_{m=-\infty}^{\infty}$Jm(k$\displaystyle\rho_{<}^{}$)Hm(1)(k$\displaystyle\rho_{\gt}^{}$)e im($\scriptstyle\phi$ - $\scriptstyle\phi$') . (2)

1.3 The magnetic Field

For the solenoid the vector potential is then

A$\scriptstyle\phi$ = $\displaystyle{\frac{i 2 \pi^2 a N I(\omega) }{c}}$J1(k$\displaystyle\rho_{<}^{}$)H1(1)(k$\displaystyle\rho_{\gt}^{}$) (3)

where $\rho_{<}^{}$ is the smaller of a and $\rho$ , and $\rho_{\gt}^{}$ is the larger. Taking curl of $\vec{A}$ , we get

Bz = $\displaystyle{\textstyle\frac{1}{\rho}}$$\displaystyle{\frac{\partial}{\partial \rho}}$($\displaystyle\rho$A$\scriptstyle\rho$) . (4)

We can use the identify for Bessel functions C1 , that

$\displaystyle{\textstyle\frac{1}{x}}$$\displaystyle{\frac{d}{dx}}$[xC1(x)] = C0(x) (5)

to give the magnetic field

Bz = $\displaystyle{\frac{i 2 \pi^2 k a N I(\omega)}{c}}$$\displaystyle\left \{
J_0(k \rho) H_1^{(1)}(ka) & \rho < a \ H_0^{(1)}(k\rho ) J_1(ka) & \rho \gt a \ \end{array}\right.$ (6)

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

1.4 Triangular Current Results

For a triangular wave current, I have
I(t) = $\displaystyle{\frac{8 I_0}{\pi^2}}$$\displaystyle\sum_{n \ \rm{odd}}^{}$$\displaystyle{\frac{\cos(n \omega_0 t)}{n^2}}$   
  = Re $\displaystyle{\frac{8 I_0}{\pi^2}}$$\displaystyle\sum_{n \ \rm{odd}}^{}$$\displaystyle{\frac{e^{i n \omega_0 t}}{n^2}}$ (7)
where 2$\pi$/$\omega$ is the period of the oscillation. Plugging this in, the magnetic field as a function of time becomes

Bz($\displaystyle\rho$,t) = Re $\displaystyle\sum_{n \ {\rm odd}}^{}$$\displaystyle{\frac{i 16 \omega_0 a N I_0}{n c^2}}$e - in$\scriptstyle\omega_{0}$t$\displaystyle\left \{
J_0(n \omega_0 \rho/c) H_1^{(1)}(n \om...
 ...{(1)}( n \omega_0 \rho /c) J_1(\omega_0 a/c) & \rho \gt a \ \end{array}\right.$ (8)

Defining x = $\rho$/a , $\beta$ = $\omega_{0}^{}$a/c , $\tau$ = $\omega_{0}^{}$t , and B0 = 4$\pi$NI0/c this can be written

Bz($\displaystyle\rho$,t) = $\displaystyle{\frac{B_0 4\beta}{\pi}}$$\displaystyle\sum_{n \ {\rm odd}}^{}$$\displaystyle{\textstyle\frac{1}{n}}$$\displaystyle\left \{
J_0(n \beta x) [ J_1(n \beta) \sin(n \...
 ... ) \sin(n \tau) - N_0(n \beta x)\cos(n \tau) ] & \rho < a \ \end{array}\right.$ (9)

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