The magnetic needle of a vibration magnetomet | vedclass

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(B) The frequency of oscillation $n$ of a vibration magnetometer is given by $n = \frac{1}{2 \pi} \sqrt{\frac{MH}{I}}$,where $M$ is the magnetic momen

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$\sqrt{63}$

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(B) The frequency of oscillation $n$ of a vibration magnetometer is given by $n = \frac{1}{2 \pi} \sqrt{\frac{MH}{I}}$,where $M$ is the magnetic moment,$H$ is the horizontal component of the earth's magnetic field,and $I$ is the moment of inertia. Thus,$n \propto \sqrt{B_{net}}$,where $B_{net}$ is the net magnetic field.Case $1$: Initially,$n_1 = 12$ oscillations/min with field $H$.Case $2$: With the bar magnet,$n_2 = 15$ oscillations/min. The net field is $H + H_1$,where $H_1$ is the field due to the bar magnet. $\frac{n_2}{n_1} = \sqrt{\frac{H + H_1}{H}} \Rightarrow \frac{15}{12} = \sqrt{1 + \frac{H_1}{H}} \Rightarrow \frac{25}{16} = 1 + \frac{H_1}{H} \Rightarrow \frac{H_1}{H} = \frac{9}{16}$.Case $3$: When poles are interchanged,the field becomes $H - H_1$. Let the new frequency be $n_3$.$\frac{n_3}{n_1} = \sqrt{\frac{H - H_1}{H}} = \sqrt{1 - \frac{H_1}{H}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.$n_3 = 12 	imes \frac{\sqrt{7}}{4} = 3\sqrt{7} = \sqrt{9 	imes 7} = \sqrt{63}$ oscillations per minute.

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