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(A) The time period of a magnetic needle oscillating in a horizontal plane is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $B_H = B \cos \delta$ i

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

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(A) The time period of a magnetic needle oscillating in a horizontal plane is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $B_H = B \cos \delta$ is the horizontal component of the Earth's magnetic field and $\delta$ is the angle of dip.Given $\delta = 60^{\circ}$,we have $T = 2\pi \sqrt{\frac{I}{M B \cos 60^{\circ}}}$.When the needle oscillates in the vertical plane coinciding with the magnetic meridian,it experiences the total magnetic field $B$. The time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I}{M B}}$.Dividing the two equations: $\frac{T'}{T} = \sqrt{\frac{M B \cos 60^{\circ}}{M B}} = \sqrt{\cos 60^{\circ}}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we get $\frac{T'}{T} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Therefore,$T' = \frac{T}{\sqrt{2}}$.

$A$ 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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