JEE Main & Advanced Physics Magnetism Vibration Magnetometer

Vibration magnetometer is used for comparison of magnetic moments and magnetic fields. This device works on the principle, that whenever a freely suspended magnet in a uniform magnetic field, is disturbed from it's equilibrium position, it starts vibrating about the mean position.

Time period of oscillation of experimental bar magnet (magnetic moment M) in earth's magnetic field \[({{B}_{H}})\] is given by the formula. \[T=2\pi \sqrt{\frac{I}{M{{B}_{H}}}}\]; where, \[I=\] moment of inertia of short bar magnet \[=\frac{w{{L}^{2}}}{12}\] (w = mass of bar magnet)

(1) Determination of magnetic moment of a magnet :  The experimental (given) magnet is put into vibration magnetometer and it's time period T is determined. Now \[T=2\pi \sqrt{\frac{I}{M{{B}_{H}}}}\,\,\Rightarrow \,M=\frac{4{{\pi }^{2}}I}{{{B}_{H}}.{{T}^{2}}}\]

(2) Comparison of horizontal components of earth's magnetic field at two places

\[T=2\pi \sqrt{\frac{I}{M{{B}_{H}}}}\] ; since I and M of the magnet are constant,

So \[{{T}^{2}}\propto \frac{1}{{{B}_{H}}}\,\,\Rightarrow \,\,\frac{{{({{B}_{H}})}_{1}}}{{{({{B}_{H}})}_{2}}}=\frac{T_{2}^{2}}{T_{1}^{2}}\]

(3) Comparison of magnetic moment of two magnets of same size and mass

\[T=2\pi \sqrt{\frac{I}{M.{{B}_{H}}}}\] ; Here I and BH are constants. 

So \[M\propto \frac{1}{{{T}^{2}}}\]  \[\Rightarrow \]\[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{T_{2}^{2}}{T_{1}^{2}}\]

(4) Comparison of magnetic moments by sum and difference method

Sum position

Net magnetic moment

\[{{M}_{s}}={{M}_{1}}+{{M}_{2}}\]

Net moment of inertia \[{{l}_{s}}={{l}_{1}}+{{l}_{2}}\]

Time period of oscillation of this pair in earth's magnetic field \[({{B}_{H}})\]

\[{{T}_{s}}=2\pi \sqrt{\frac{{{I}_{s}}}{{{M}_{s}}\,{{B}_{H}}}}\]\[=2\pi \sqrt{\frac{{{I}_{1}}+{{I}_{2}}}{({{M}_{1}}+{{M}_{2}}){{B}_{H}}}}\]              ....(i)

Frequency  \[{{\nu }_{s}}=\frac{1}{2\pi }\sqrt{\frac{({{M}_{1}}+{{M}_{2}}){{B}_{H}}}{{{I}_{s}}}}\]

Difference position  

Net magnetic moment

\[{{M}_{d}}={{M}_{1}}+{{M}_{2}}\]

Net moment of inertia \[{{I}_{d}}={{I}_{1}}+{{I}_{2}}\]

and  \[{{T}_{d}}=2\pi \sqrt{\frac{{{I}_{d}}}{{{M}_{d}}\,{{B}_{H}}}}\]\[=2\pi \sqrt{\frac{{{I}_{1}}+{{I}_{2}}}{({{M}_{1}}-{{M}_{2}}){{B}_{H}}}}\]   ....(ii)

and \[{{\nu }_{d}}=\frac{1}{2\pi }\sqrt{\frac{({{M}_{1}}+{{M}_{2}})\,{{B}_{H}}}{({{I}_{1}}+{{I}_{2}})}}\]. From equation (i) and (ii) we get

\[\frac{{{T}_{s}}}{{{T}_{d}}}=\sqrt{\frac{{{M}_{1}}-{{M}_{2}}}{{{M}_{1}}+{{M}_{2}}}}\]\[\Rightarrow \]\[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{T_{d}^{2}+T_{s}^{2}}{T_{d}^{2}-T_{s}^{2}}=\frac{\nu _{s}^{2}+\nu _{d}^{2}}{\nu _{s}^{2}-\nu _{d}^{2}}\]

(5) To find the ratio of magnetic field : Suppose it is required to find the ratio \[\frac{B}{{{B}_{H}}}\] where B is the field created by magnet and \[{{B}_{H}}\] is the horizontal component of earth's magnetic field.

To determine \[\frac{B}{{{B}_{H}}}\] a primary (main) magnet is made to first oscillate in earth's magnetic field \[({{B}_{H}})\] alone and it's time period of oscillation (T) is noted.

\[T=2\pi \sqrt{\frac{I}{M\,{{B}_{H}}}}\]

and frequency \[\nu =\frac{1}{2\pi }\sqrt{\frac{M\,{{B}_{H}}}{I}}\]

Now a secondary magnet placed near the primary magnet so primary magnet oscillate in a new field with is the resultant of B and BH and now time period, is noted again.

\[T'=2\pi \sqrt{\frac{I}{M(B+{{B}_{H}})}}\]

or \[\nu \,'=\frac{1}{2\pi }\sqrt{\frac{M(B+{{B}_{H}})}{I}}\]

\[\Rightarrow \]\[\frac{B}{{{B}_{H}}}={{\left( \frac{\nu \,'}{\nu } \right)}^{2}}-1\]