which satisfies (4.1) when and . Thus, is a solution to . This is equivalent to

In addition, and so, taking the curl of (4.12) we obtain

Using a vector identity we obtain

Thus, as the first term in the middle expression is zero, we have

For , (4.12) and (4.13) become

and

respectively. To solve these equations we note that they are linear equations and that the coefficients are constants and so we look for seperable solutions of the form

Substituting into (4.15) gives

Dividing by and rearranging gives

Taking the constant as we obtain equations for and as

These are easily solved in terms of trigonometric functions and expontential functions respectively to give one possible solution as

and

The field lines are shown in Figure 4.5.

- Potential Problems leading to Laplace's Equation
- Non-potential current flowing inside a cylindrical magnetic loop