In a vibration magnetometer, time period of a bar magnet oscillating in horizontal component of eart

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5.Magnetism and Matter

medium

In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth's magnetic field is $2$ $sec$. When a magnet is brought near and parallel to it, the time period reduces to $1 $ $sec$. The ratio $H/F $ of the horizontal component $H$ and the field $F$ due to magnet will be

$3$

$1/3$

$\sqrt 3 $

$1/\sqrt 3 $

Solution

(b)$T \propto \frac{1}{{\sqrt H }} \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{H_2}}}{{{H_1}}}} \Rightarrow \;\frac{2}{1} = \sqrt {\frac{{H + F}}{H}} \Rightarrow F = 3H$
or $\frac{H}{F} = \frac{1}{3}$

Standard 12

Physics

Vibration magnetometer is used for comparing

easy

Two bar magnets of the same mass, length and breadth but magnetic moments $M$ and $2\,M$ respectively, when placed in same position, time period is $3 \,sec$. What will be the time period when they are placed in different position

hard

A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $4\, sec$. If one of the magnets is taken away, find the period of oscillation of the other in the same field (in $sec.$) :-

hard

Keeping dissimilar poles of two magnets of equal pole strength and length same side, their time period will be