Question

To compare magnetic moments of two magnets by vibration magnetometer, 'sum and difference method' is better because
A. Determination of moment of inertia is not needed which minimizes the errors
B. Less observations are required
C. Comparatively less calculations
D. All the above

Answer 1

Hint:Magnetic moment is the measure of the strength or ability of any magnet to align itself in the direction of an external magnet field. Vibration magnetometer is an instrument which helps in determining the magnetic moment and also magnetic field. In this question we need to compare the magnetic moment of two magnets. The magnetic moment of the first magnet is \[{{M}_{1}}\] and of the other magnet it is \[{{M}_{2}}\]. In sum and difference method, both the magnetic moment are added once and once they are subtracted from each other such as \[{{M}_{1}}+{{M}_{2}}\] and \[{{M}_{1}}-{{M}_{2}}\].

Formula used:
Using the magnetic moments obtained by the sum and difference method, we need to calculate the time period of both magnetic moments. The general formula for time period is given as,
\[T=2\pi \sqrt{\dfrac{I}{M{{B}_{H}}}}\]
Where, I is moment of inertia, M is magnetic moment, B is external magnetic field.

Complete step by step solution:
The time period (time taken to complete one oscillation) for magnetic moment \[{{M}_{1}}+{{M}_{2}}\] (obtained by vibration magnetometer) is given as,
\[{{T}_{1}}=2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{({{M}_{1}}+{{M}_{2}}){{B}_{H}}}}\]
As the magnetic moment of both magnets is added so the moment of inertia which will be different for both magnets (I1 and I2) will also get added. Also both magnets are inserted within the same uniform magnetic field B, so the uniform magnetic field will not change for both time periods.

The time period for magnetic moment \[{{M}_{1}}-{{M}_{2}}\] (as per difference method) is given as,
\[{{T}_{2}}=2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{({{M}_{1}}-{{M}_{2}}){{B}_{H}}}}\]
Now divide the time period of magnetic moment \[{{M}_{1}}+{{M}_{2}}\], \[{{T}_{1}}\] to the time period of magnetic moment \[{{M}_{1}}-{{M}_{2}}\], \[{{T}_{2}}\] to find the relation between both magnetic moments such as
\[\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{({{M}_{1}}+{{M}_{2}}){{B}_{H}}}}}{2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{({{M}_{1}}-{{M}_{2}}){{B}_{H}}}}}\]

Now, as we divide the intensity which is the same will cancel out, external magnetic field will also vanish and also 2 pie. Then we will left with only magnetic moment calculated with the help of sum and difference method such as
\[\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{\sqrt{\dfrac{1}{({{M}_{1}}+{{M}_{2}})}}}{\sqrt{\dfrac{1}{({{M}_{1}}-{{M}_{2}})}}} \\ \]
\[\therefore \dfrac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\dfrac{{{M}_{1}}-{{M}_{2}}}{{{M}_{1}}+{{M}_{2}}}}\]
In this way we can say, using the sum and difference method we eliminate the need to calculate moment of inertia. Also we only need to calculate magnetic moment which involves no calculation of magnetic field and constant and so we need less observation and calculations.

Hence, option D is correct.

Note: Vibrational magnetometer work on the principle of Faraday’s law of induction. Actually when a magnet is placed in the vibrational magnetometer it starts to vibrate at once disturbance in the uniform magnetic field (induce magnetic field). Now more the magnet oscillate in less time (less time period) means it is aligning itself with external field easily and effectively and its magnetic strength (magnetic moment) is large (that’s why inverse relationship between time period and magnetic moment)