\(\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{\mathrm{B}_{1} \cos 45^{\circ}}{\mathrm{B}_{2} \cos 30^{\circ}}\)
\({{\text{f}}_2} = \frac{1}{{2\pi }}\sqrt {\frac{{\mu {{\text{B}}_2}\cos {{30}^\circ }}}{1}} \)
\(\therefore \frac{\mathrm{B}_{1}}{\mathrm{B}_{2}}= 0.7\)