We rewrite Eqs.(2.1-45) and (2.1-60) in normalized form for the derivation of the associated magnetic field strength. First we get with the help of Eqs.(2.1-32) and (2.1-48):
$$\begin{array}{*{20}{c}} {\frac{{{{\tau }_{{1m}}}{{\tau }_{{2m}}}}}{{{{\tau }_{{2m}}} - {{\tau }_{{1m}}}}} = - \frac{{{{\tau }_{{mp}}}}}{{{{{\left( {1 - 8\alpha '{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}} \\ {{{\tau }_{{1m}}} = {{\tau }_{{mp}}}\frac{{1 + {{{\left( {1 - 8{{{\alpha '}}_{1}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}{{4{{{\alpha '}}_{1}}{{\tau }_{{mp}}}}},{{\tau }_{{2m}}} = {{\tau }_{{mp}}}\frac{{1 + {{{\left( {1 - 8{{{\alpha '}}_{1}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}{{4{{{\alpha '}}_{1}}{{\tau }_{{mp}}}}}} \\ {t/{{\tau }_{{1m}}} = {{{v'}}_{1}}\theta /pq,t/{{\tau }_{{2m}}} = {{{v'}}_{2}}\theta /pq,\theta = t/\tau ,p = {{\tau }_{{mp}}}/{{\tau }_{p}},q = {{\tau }_{p}}/\tau } \\ {{{{v'}}_{1}} = \frac{{4{{{\alpha '}}_{1}}{{\tau }_{{mp}}}}}{{1 + {{{\left( {1 - 8{{{\alpha '}}_{1}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}},{{{v'}}_{2}} = \frac{{4{{{\alpha '}}_{1}}{{\tau }_{{mp}}}}}{{1 - {{{\left( {1 - 8{{{\alpha '}}_{1}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}} \\ {\frac{{{{\tau }_{{3m}}}{{\tau }_{{4m}}}}}{{{{\tau }_{{3m}}} - {{\tau }_{{4m}}}}} = - \frac{{{{\tau }_{{mp}}}}}{{{{{\left( {1 - 8{{{\alpha '}}_{2}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}} \\ {{{\tau }_{{3m}}} = {{\tau }_{{mp}}}\frac{{1 + {{{\left( {1 - 8{{{\alpha '}}_{2}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}{{4{{{\alpha '}}_{2}}{{\tau }_{{mp}}}}},{{\tau }_{{4m}}} = {{\tau }_{{mp}}}\frac{{1 + {{{\left( {1 - 8{{\alpha }_{2}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}{{4{{\alpha }_{2}}{{\tau }_{{mp}}}}}} \\ {t/{{\tau }_{{3m}}} = {{v}_{3}}\theta /pq,t/{{\tau }_{{4m}}} = {{v}_{4}}\theta /pq,\theta = t/\tau ,p = {{\tau }_{{mp}}}/{{\tau }_{p}},q = {{\tau }_{p}}/\tau } \\ {{{v}_{3}} = \frac{{4{{\alpha }_{2}}{{\tau }_{{mp}}}}}{{1 + {{{\left( {1 - 8{{\alpha }_{2}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}},{{{v'}}_{4}} = \frac{{4{{\alpha }_{2}}{{\tau }_{{mp}}}}}{{1 - {{{\left( {1 - 8{{\alpha }_{2}}{{\tau }_{{mp}}}} \right)}}^{{{{1} \left/ {2} \right.}}}}}}} \\ \end{array}$$
((1))
