An improvement in the calculation of the magnetic field for an arbitrary geometry coil with rectangular cross section Int. J. Numer. Model . 2005; 18 (6):493–504

The above article was published in Volume 18, Issue 6, Pages 493–504, (October 2005). A number of imperfections were subsequently identified and have been corrected below. On page 497, the following equation has been corrected to:\documentclass{article} \footskip=0pc\pagestyle{empty} \begin{document} \begin{eqnarray*} S_{xn} & = & \left\{ t\,\sinh^{-1}\,\frac{t\,\sin(\theta) + R\,\cos(\theta)}{\cos(\theta) \sqrt{t^{2} + Q^{2}}} + R\,\cos(\theta)\,\sinh^{-1}\,\frac{t + R\,\sin(\theta)\,\cos (\theta)}{\cos(\theta)\sqrt{R^{2}\,\cos^{2}(\theta) + Q^{2}}}\right.\\ && \left. \left. + Q\,\tan^{-1}\,\frac{Q^{2}\,\sin(\theta) - tR\,\cos(\theta)}{Q\sqrt{t^{2} + 2Rt\,\sin(\theta)\,\cos(\theta) + (R^{2} + Q^{2})\,\cos^{2}(\theta)}}\right\} \right|_{t = L_{1}}^{t = L_{2}} \end{eqnarray*}\end{document}On page 501, the following equation has been corrected to:\documentclass{article} \footskip=0pc\pagestyle{empty} \begin{document} \begin{eqnarray*} I_{x1} & = & \int\nolimits_{-d}^{d} I_{y1}\,{\rm{d}} x' = \left[ t\,\sinh^{-1} \,\frac{t\,\sin(\theta) + R\,\cos(\theta)}{\cos(\theta)\sqrt{t^{2} + Q^{2}}} + R\,\cos(\theta)\,\sinh^{-1}\,\frac{t + R\,\sin(\theta)\,\cos(\theta)} {\cos(\theta)\sqrt{R^{2}\,\cos^{2}(\theta) + Q^{2}}}\right.\\ && \left. \left. + Q\,\tan^{-1}\,\frac{Q^{2}\,\sin(\theta) - tR\,\cos(\theta)}{Q\sqrt{t^{2} + 2Rt\,\sin(\theta)\,\cos(\theta) + (R^{2} + Q^{2})\,\cos^{2}(\theta)}} \right]\right|_{t = L_{1}}^{t = L_{2}} \end{eqnarray*}\end{document}On page 502, the following equation has been corrected to:\documentclass{article} \footskip=0pc\pagestyle{empty} \begin{document} \begin{eqnarray*} I_{z2} & = & \int\nolimits_{- c}^{c} I_{x2}\,{\rm{d}} x' = \left\{ t\,\sin(\theta) \,\sinh^{-1}\,\frac{L + R\,\sin(\theta)\,\cos(\theta)}{\cos(\theta) \sqrt{t^{2} + R^{2}\,\cos^{2}(\theta)}} - t\,\sinh^{-1}\, \frac{L\,\sin(\theta) + R\,\cos(\theta)}{\cos(\theta) \sqrt{t^{2} + L^{2}}}\right. \\ && - R\,\cos^{2}(\theta)\,\sinh^{-1}\,\frac{t\,\cos(\theta)}{\sqrt{L^{2} + 2LR\,\sin(\theta)\,\cos(\theta) + R^{2}\,\cos^{2}(\theta)}}\\ && - R\,\sin(\theta)\,\cos(\theta)\,\tan^{-1}\, \frac{t(L + R\,\sin(\theta)\,\cos(\theta))} {R\,\cos(\theta)\sqrt{t^{2}\,\cos^{2}(\theta) + L^{2} + 2LR\,\sin(\theta) \,\cos(\theta) + R^{2}\,\cos^{2}(\theta)}}\\ && \left. \left. + L\,\tan^{-1}\, \frac{t(L\,\sin(\theta) + R\,\cos(\theta))}{L\sqrt{t^{2}\,\cos^{2}(\theta) + L^{2} + 2LR\,\sin(\theta)\,\cos(\theta) + R^{2}\,\cos^{2}(\theta)}} \right\} \right|_{t = Q_{1}}^{t = Q_{2}} \end{eqnarray*}\end{document}