A magnetic compass needle oscillates 30 times | vedclass

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Question

$\mathrm{f}_{1}=\frac{1}{2 \pi} \sqrt{\frac{\mu \mathrm{B}_{1} \cos 45^{\circ}}{1}} \quad$

$\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{\mathrm{B}_

Vedclass · Accepted Answer

$0.7$

Vedclass · Answer

$\mathrm{f}_{1}=\frac{1}{2 \pi} \sqrt{\frac{\mu \mathrm{B}_{1} \cos 45^{\circ}}{1}} \quad$

$\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{\mathrm{B}_{1} \cos 45^{\circ}}{\mathrm{B}_{2} \cos 30^{\circ}}$

${{	ext{f}}_2} = \frac{1}{{2\pi }}\sqrt {\frac{{\mu {{	ext{B}}_2}\cos {{30}^\circ }}}{1}} $

$	herefore \frac{\mathrm{B}_{1}}{\mathrm{B}_{2}}= 0.7$

A magnetic compass needle oscillates $30$ times per minute at a place where the dip is $45^o$, and $40$ times per minute where the dip is $30^o$. If $B_1$ and $B_2$ are respectively the total magnetic field due to the earth at the two places, then the ratio $B_1/B_2$ is best given by

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