Taking curl of the equation \nabla\times\mathbf{H}=\lambda\mathbf{H} and using this same equation, we get :\nabla\times(\nabla\times\mathbf{H}) = \lambda^2\mathbf{H}. In the vector identity \nabla \times \left( \nabla \times \mathbf{H} \right) = \nabla(\nabla \cdot \mathbf{H}) - \nabla^{2}\mathbf{H}, we can set \nabla\cdot\mathbf{H}=0 since it is solenoidal, which leads to a vector
Helmholtz equation, :\nabla^2\mathbf{H}+\lambda^2\mathbf{H}=0. Every solution of above equation is not the solution of original equation, but the converse is true. If \psi is a scalar function which satisfies the equation \nabla^2\psi + \lambda^2\psi=0, then the three
linearly independent solutions of the vector
Helmholtz equation are given by :\mathbf{L} = \nabla\psi,\quad \mathbf{T} = \nabla\times\psi\mathbf{\hat n}, \quad \mathbf{S} = \frac{1}{\lambda}\nabla\times\mathbf{T} where \mathbf{\hat n} is a fixed unit vector. Since \nabla\times\mathbf{S} =\lambda\mathbf{T}, it can be found that \nabla\times(\mathbf{S}+\mathbf{T})=\lambda(\mathbf{S}+\mathbf{T}). But this is same as the original equation, therefore \mathbf{H}=\mathbf{S}+\mathbf{T}, where \mathbf{S} is the poloidal field and \mathbf{T} is the toroidal field. Thus, substituting \mathbf{T} in \mathbf{S}, we get the most general solution as :\mathbf{H} = \frac{1}{\lambda}\nabla\times(\nabla\times\psi\mathbf{\hat n}) + \nabla \times \psi \mathbf{\hat n}. ==Cylindrical polar coordinates==