Question

The time period of oscillation of magnet in a vibrating magnetometer is 1.5 sec. The time period of oscillation of another magnet similar in size and mass but having one-fourth the magnetic moment than that of the first magnet oscillating at the same place will be
A. 0.75 sec
B. 1.5 sec
C. 3.0 sec
D. 6.0 sec

Answer 1

Hint: In this question, the time period (\[{{T}_{1}}\]) of oscillation of one magnet in the vibrating magnetometer is given, 1.5 sec. It means the taken magnet is completing one oscillation in 1.5 sec. In other cases, we have taken the other magnet which is similar in mass and size to the first magnet but has different magnetic properties (magnetic moment). Magnetic moment (tendency to align itself with the external magnetic field) of the second magnet is equal to one fourth of the magnetic moment of the first magnet. Let \[{{M}_{1}}\] is magnetic moment of first magnet, \[{{M}_{2}}\] is the magnetic moment of second magnet which is equal to \[1/4\] (\[{{M}_{1}}\]). So, for the second magnet we need to find the time period (\[{{T}_{2}}\]).

Formula used:
The general formula of time period is given as
\[T=2\pi \sqrt{\dfrac{I}{M{{B}_{H}}}}\]
Where I is the moment of inertia, M is magnetic moment and B is external magnetic field.

Complete step by step solution:
Vibrating magnetometer is an instrument which is used to determine the magnetic moment of the placed magnet in that instrument. This instrument is the source of a uniform magnetic field. We have taken two magnets of the same size and mass so, we can say both magnets are the same.

When we introduce the first magnet in that instrument, it experiences a magnetic field and started to oscillate on once disturbance and how much it align itself with the external field is its magnetic moment, let say it is \[{{M}_{1}}\]. By using the values of external magnetic field, magnetic moment and moment of inertia, calculated time period, \[{{T}_{1}}\] (one oscillation in what time) is given, which is 1.5sec such as,
\[{{T}_{1}}=2\pi \sqrt{\dfrac{I}{{{M}_{1}}{{B}_{H}}}}=1.5\sec \]

Now, when we introduce second magnet of same size and mass in vibrating magnetometer, observed value of magnetic moment (\[{{M}_{2}}\]) was one fourth of magnetic moment (\[{{M}_{1}}/4\]) of first magnet . So the time period (\[{{T}_{2}}\]) for magnetic moment, \[{{M}_{2}}\] is given as,

\[{{T}_{2}}=2\pi \sqrt{\dfrac{I}{{{M}_{2}}{{B}_{H}}}}\]
\[\Rightarrow {{T}_{2}}=2\pi \sqrt{\dfrac{I}{\dfrac{{{M}_{1}}}{4}{{B}_{H}}}}\]
\[\Rightarrow {{T}_{2}}=2\pi \sqrt{\dfrac{I\times 4}{{{M}_{1}}{{B}_{H}}}}\]
\[\Rightarrow {{T}_{2}}=2\times 2\pi \sqrt{\dfrac{I}{{{M}_{1}}{{B}_{H}}}}\]
As we can noticed that that the term after 2 is same to time period of first magnet such as
\[{{T}_{2}}=2\times {{T}_{1}}\]
As the value of time period (\[{{T}_{1}}\]) of first magnet is given in question, 1.5 sec so,
\[{{T}_{2}}=2\times 1.5\]
\[\therefore {{T}_{2}}=3\sec \]
Therefore, the second magnet completes one oscillation in 3 sec.

Hence, option C is correct.

Note:As according to the question we need to take two magnets of same mass and size thus the moment of inertia for the time period of both magnets will be the same and we took it as I. Also, the magnetic field experienced by both magnets is also same and we taken it as B. we can get same result by also dividing the time period of both the magnets such as
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}} \\ $
$\Rightarrow \dfrac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\dfrac{{{M}_{1}}/4}{{{M}_{1}}}} \\ $
$\Rightarrow \dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{1}{2} \\ $
$\Rightarrow \dfrac{1.5}{{{T}_{2}}}=\dfrac{1}{2} \\ $
$\therefore {{T}_{2}}=3$