A compass needle of oscillation magnetometer | vedclass

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Question

$T =2 \pi \sqrt{\frac{ I }{ B _{ H } M }}$

$T _{1}=3\,sec =2 \pi \sqrt{\frac{ I }{\left( B _{ p } \cos 30^{\circ}\right) M }}$

$T _{2}=6\,sec =2

Vedclass · Accepted Answer

$\sqrt{3}: 4$

Vedclass · Answer

$T =2 \pi \sqrt{\frac{ I }{ B _{ H } M }}$

$T _{1}=3\,sec =2 \pi \sqrt{\frac{ I }{\left( B _{ p } \cos 30^{\circ}\right) M }}$

$T _{2}=6\,sec =2 \pi \sqrt{\frac{ I }{\left( B _{ Q } \cos 60^{\circ}\right) M }}$

$\frac{3}{6}=\sqrt{\frac{ B _{ Q } / 2}{B _{ p } \frac{\sqrt{3}}{2}}}$

$\frac{3}{6}=\sqrt{\frac{ B _{ Q }}{\sqrt{3} B _{ p }}}$

$\frac{ B _{ Q }}{\frac{3}{4}}=\frac{ B _{ p }}{ B _{ Q }}=\sqrt{3}: 4$

A compass needle of oscillation magnetometer oscillates $20$ times per minute at a place $P$ of dip $30^{\circ}$. The number of oscillations per minute become $10$ at another place $Q$ of $60^{\circ} dip$. The ratio of the total magnetic field at the two places $\left(B_{Q}: B_{p}\right)$ is.

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