A magnetic needle oscillates in a horizontal | vedclass

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Question

(a)

$T=2 \pi \sqrt{\frac{1}{M B \cos 60^{\circ}}} \Rightarrow \frac{T^{\prime}}{T}=\sqrt{\cos 60^{\circ}}$

$T^{\prime}=2 \pi \sqrt{\frac{I}{M B}

Vedclass · Accepted Answer

$\frac{T}{\sqrt{2}}$

Vedclass · Answer

(a)

$T=2 \pi \sqrt{\frac{1}{M B \cos 60^{\circ}}} \Rightarrow \frac{T^{\prime}}{T}=\sqrt{\cos 60^{\circ}}$

$T^{\prime}=2 \pi \sqrt{\frac{I}{M B}} \Rightarrow \frac{T^{\prime}}{T}=\frac{1}{\sqrt{2}} \Rightarrow T^{\prime}=\frac{T}{\sqrt{2}}$

A magnetic needle oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^{\circ}$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

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