Question

Two magnets A and B are identical in mass, length and breadth but have different magnetic moments. In a vibration magnetometer, if the time period of B is twice the time period of A. The ratio of the magnetic moments MA​/MB​ of the magnets will be
A. $\dfrac{1}{2}$
B. 2
C. 4
D. $\dfrac{1}{4}$

Answer 1

Hint: We have taken two identical magnets (A and B whose length, breath, and mass is same but different tendency to align itself with external field , different magnetic moment let say \[{{M}_{A}}\] (A magnet) and \[{{M}_{B}}\] (B magnet). We need to find the ratio of magnetic moment of A magnet to the magnetic moment of B magnet such as \[{{M}_{A}}/{{M}_{B}}\]. With increase of magnetic moment, alignment increases with external magnetic field and time to complete one oscillation decreases (Time period decreases). Inverse relationship between magnetic moment and time period of magnet.

Formula used:
The time period of oscillation of any magnet in the applied magnetic field is given as
\[T=2\pi \sqrt{\dfrac{I}{M{{B}_{H}}}}\]
Where, $I=\dfrac{m(l^2+b^2)}{12}$ is the moment of inertia of the vibrating magnet,
m = mass of magnet
l = Length of magnet
b = breadth of magnet
As we can notice the inverse relationship between time period and magnetic moment of any magnet such as,
$T\propto \dfrac{1}{\sqrt{M}}$

Complete step by step solution:
 The vibrational magnetometer serves as a tool for comparing the magnetic moments and magnetic fields of two magnets. The vibration magnetometer in the magnetic meridian is used to measure the duration of a bar magnet's oscillation. When we place both the magnets in an instrument, vibrational magnetometer, both magnets start to oscillate or vibrate in the external field at once disturbance and thus, we get the magnetic moment of both magnets through time period.

The magnetic moment of A magnet is \[{{M}_{A}}\] and the magnetic moment of B magnet is \[{{M}_{B}}\] then the time period of both A and B (\[{{T}_{A}}\] ) and (\[{{T}_{B}}\]) magnet is given as,
\[{{T}_{A}}=2\pi \sqrt{\dfrac{I}{{{M}_{A}}{{B}_{H}}}}\]
And,
\[{{T}_{B}}=2\pi \sqrt{\dfrac{I}{{{M}_{B}}{{B}_{H}}}}\]
Or, we can write it as
${{T}_{A}}\propto \dfrac{1}{\sqrt{{{M}_{A}}}}$
And, ${{T}_{B}}\propto \dfrac{1}{\sqrt{{{M}_{B}}}}$
As we need to find the ratio of \[{{M}_{A}}/{{M}_{B}}\] so, we need to find the relation in term of magnetic moment such as,
$\sqrt{{{M}_{A}}}\propto \dfrac{1}{{{T}_{A}}}$
$\Rightarrow {{M}_{A}}\propto \dfrac{1}{{{({{T}_{A}})}^{2}}}$............(squaring on both side)
And,
$\sqrt{{{M}_{B}}}\propto \dfrac{1}{{{T}_{B}}}$
$\Rightarrow {{M}_{B}}\propto \dfrac{1}{{{({{T}_{B}})}^{2}}}$.......(by squaring both side)

Finding ratio $M_A/M_B$
$\Rightarrow \dfrac{{{M}_{A}}}{{{M}_{B}}}=\dfrac{\dfrac{1}{{{T}_{A}}^{2}}}{\dfrac{1}{{{T}_{B}}^{2}}}$
$\Rightarrow \dfrac{{{M}_{A}}}{{{M}_{B}}}=\dfrac{T_{B}^{2}}{T_{A}^{2}}$
As in given question, the time period of B (\[{{T}_{B}}\]) is twice of time period of A (\[{{T}_{A}}\]) (\[{{T}_{B}}\text{ }=\text{ }2{{T}_{A}}\] ) such as
$\Rightarrow \dfrac{{{M}_{A}}}{{{M}_{B}}}=\dfrac{{{(2{{T}_{A}})}^{2}}}{T_{A}^{2}}$
$\Rightarrow \dfrac{{{M}_{A}}}{{{M}_{B}}}=\dfrac{4{{T}_{A}}^{2}}{T_{A}^{2}}$
$\therefore \dfrac{{{M}_{A}}}{{{M}_{B}}}=4$

Thus, the correct option is C.

Note: In this question both magnets are placed in the vibrational magnetometer and experience the uniform magnetic field so that the strength of the magnet (magnetic moment) can be determined with respect to each other. Thus the magnetic field B is same for both magnets and also both magnets are of same size and mass due to moment of inertia I is also the same of both magnets. We can say both magnetic field B and moment of inertia I will be treated as constant in the given question conditions.