A small bar magnet A oscillates in a horizont | vedclass

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(A) The period of oscillation of a bar magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is

Vedclass · Accepted Answer

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

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(A) The period of oscillation of a bar magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the effective magnetic field.In the horizontal plane,the effective field is the horizontal component of the Earth's magnetic field,$B_H = B_e \cos \phi$,where $B_e$ is the total magnetic field and $\phi$ is the angle of dip.Thus,$T = 2\pi \sqrt{\frac{I}{M B_H}}$.When oscillating in the vertical plane coinciding with the magnetic meridian,the effective field is the total magnetic field $B_e$.Let the new period be $T'$. Then $T' = 2\pi \sqrt{\frac{I}{M B_e}}$.Taking the ratio: $\frac{T'}{T} = \sqrt{\frac{B_H}{B_e}} = \sqrt{\cos \phi}$.Given $\phi = 60^o$,we have $\frac{T'}{T} = \sqrt{\cos 60^o} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Therefore,$T' = \frac{T}{\sqrt{2}}$.

$A$ small bar magnet $A$ oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^o$. 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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