Taking curl of the equation ∇ × H = λ H {\displaystyle \nabla \times \mathbf {H} =\lambda \mathbf {H} } and using this same equation, we get
In the vector identity ∇ × ( ∇ × H ) = ∇ ( ∇ ⋅ H ) − ∇ 2 H {\displaystyle \nabla \times \left(\nabla \times \mathbf {H} \right)=\nabla (\nabla \cdot \mathbf {H} )-\nabla ^{2}\mathbf {H} } , we can set ∇ ⋅ H = 0 {\displaystyle \nabla \cdot \mathbf {H} =0} since it is solenoidal, which leads to a vector Helmholtz equation,
Every solution of above equation is not the solution of original equation, but the converse is true. If ψ {\displaystyle \psi } is a scalar function which satisfies the equation ∇ 2 ψ + λ 2 ψ = 0 {\displaystyle \nabla ^{2}\psi +\lambda ^{2}\psi =0} , then the three linearly independent solutions of the vector Helmholtz equation are given by
where n ^ {\displaystyle \mathbf {\hat {n}} } is a fixed unit vector. Since ∇ × S = λ T {\displaystyle \nabla \times \mathbf {S} =\lambda \mathbf {T} } , it can be found that ∇ × ( S + T ) = λ ( S + T ) {\displaystyle \nabla \times (\mathbf {S} +\mathbf {T} )=\lambda (\mathbf {S} +\mathbf {T} )} . But this is same as the original equation, therefore H = S + T {\displaystyle \mathbf {H} =\mathbf {S} +\mathbf {T} } , where S {\displaystyle \mathbf {S} } is the poloidal field and T {\displaystyle \mathbf {T} } is the toroidal field. Thus, substituting T {\displaystyle \mathbf {T} } in S {\displaystyle \mathbf {S} } , we get the most general solution as
Taking the unit vector in the z {\displaystyle z} direction, i.e., n ^ = e z {\displaystyle \mathbf {\hat {n}} =\mathbf {e} _{z}} , with a periodicity L {\displaystyle L} in the z {\displaystyle z} direction with vanishing boundary conditions at r = a {\displaystyle r=a} , the solution is given by34
where J m {\displaystyle J_{m}} is the Bessel function, k = ± 2 π n / L , n = 0 , 1 , 2 , … {\displaystyle k=\pm 2\pi n/L,\ n=0,1,2,\ldots } , the integers m = 0 , ± 1 , ± 2 , … {\displaystyle m=0,\pm 1,\pm 2,\ldots } and μ j {\displaystyle \mu _{j}} is determined by the boundary condition a k μ j J m ′ ( μ j a ) + m λ J m ( μ j a ) = 0. {\displaystyle ak\mu _{j}J_{m}'(\mu _{j}a)+m\lambda J_{m}(\mu _{j}a)=0.} The eigenvalues for m = n = 0 {\displaystyle m=n=0} has to be dealt separately. Since here n ^ = e z {\displaystyle \mathbf {\hat {n}} =\mathbf {e} _{z}} , we can think of z {\displaystyle z} direction to be toroidal and θ {\displaystyle \theta } direction to be poloidal, consistent with the convention.
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