Question

Two small magnets have their masses and lengths in the ratio \[1:2\]. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations, the ratio of their time period is:
A. \[\dfrac{1}{{2\sqrt 2 }}\]
B. \[\dfrac{1}{{\sqrt 2 }}\]
C. \[\left( {\dfrac{1}{2}} \right)\]
D. \[2\sqrt 2 \]

Answer 1

Hint: \[I = \dfrac{{m{l^2}}}{{12}}\], to find moment of inertia and
\[T = 2\pi \sqrt {\dfrac{I}{{MB}}} \] to find the time period.

Complete step by step answer:
We are given two small bar magnets; whose masses and their length are both in the ratio \[1:2\] .
Let us take the mass of the first bar magnet be \[{m_1}\] .
Length of the first bar magnet be \[{l_1}\] .
Mass of the second bar magnet be \[{m_2}\] .
Length of the second bar magnet \[{l_2}\] .

Let us take the ratio of the masses of the two bar magnets as follows:
\[
\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{2} \\
\dfrac{{{m_1}}}{1} = \dfrac{{{m_2}}}{2} \\
 \]
Let us take each ratio to be equal to a factor \[m\] .
So,
\[\dfrac{{{m_1}}}{1} = \dfrac{{{m_2}}}{2} = m\]
We can write:
\[{m_1} = m\] and \[{m_2} = 2m\]

Let us take the ratio of the lengths of the two bar magnets as follows:
\[
\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{1}{2} \\
\dfrac{{{l_1}}}{1} = \dfrac{{{l_2}}}{2} \\
 \]
Let us take each ratio to be equal to a factor \[l\] .
So,
\[\dfrac{{{l_1}}}{1} = \dfrac{{{l_2}}}{2} = l\]
We can write:
\[{l_1} = l\] and \[{l_2} = 2l\]

Formula which gives the moment of inertia for the first bar magnet is given by:
\[{I_1} = \dfrac{{{m_1}l_1^2}}{{12}}\]
\[{I_1} = \dfrac{{m{l^2}}}{{12}}\] …… (1)

The moment of inertia of the second magnet is given by:
\[{I_2} = \dfrac{{{m_2}l_2^2}}{{12}}\]
\[{I_2} = \dfrac{{2m \times {{\left( {2l} \right)}^2}}}{{12}}\]
\[{I_2} = \dfrac{{8m{l^2}}}{{12}}\] …… (2)

The expression which gives the maximum torque on a bar magnet is given by the expression:
\[\tau = M \times B\]
Where,
\[\tau \] indicates torque.
\[M\] indicates magnetic moment.
\[B\] indicates magnetic field.

We are given in the question that the torque experienced by the two magnets are equal, so we can write:
\[{\tau _1} = {\tau _2}\]
\[\therefore {M_1}{B_1} = {M_2}{B_2}\] …… (3)

The expression which gives the time period of oscillation of a bar magnet is given by:
\[T = 2\pi \sqrt {\dfrac{I}{{MB}}} \]
Where,
\[T\] indicates time period.

So, the time period of oscillation for the first magnet is:
\[{T_1} = 2\pi \sqrt {\dfrac{{{I_1}}}{{{M_1}{B_1}}}} \]
The time period of oscillation for the second magnet is:
\[{T_2} = 2\pi \sqrt {\dfrac{{{I_2}}}{{{M_2}{B_2}}}} \] …… (4)

Substitute, \[{M_1}{B_1} = {M_2}{B_2}\] in equation (4):
\[{T_2} = 2\pi \sqrt {\dfrac{{{I_2}}}{{{M_1}{B_1}}}} \]

Now, we find the ratio of the time period of the two magnets:

\[
\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{2\pi \sqrt {\dfrac{{{I_1}}}{{{M_1}{B_1}}}} }}{{2\pi \sqrt {\dfrac{{{I_2}}}{{{M_1}{B_1}}}} }} \\
   = \sqrt {\dfrac{{{I_1}}}{{{M_1}{B_1}}}} \times \sqrt {\dfrac{{{M_1}{B_1}}}{{{I_2}}}} \\
   = \sqrt {\dfrac{{{I_1}}}{{{I_2}}}} \\
   = \sqrt {\dfrac{{\left( {\dfrac{{m{l^2}}}{{12}}} \right)}}{{\left( {\dfrac{{8m{l^2}}}{{12}}} \right)}}} \\
 \]
Again, we simplify the above:
\[
\dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{1}{8}} \\
\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{{2\sqrt 2 }} \\
 \]

So, the correct answer is “Option A”.

Note:
In this problem we are asked to find the ratio of the time period of oscillations. It is important to note that the ratio of mass and length is not \[1:2\], rather the ratio of masses and lengths of each magnet is \[1:2\]. Take the torque of the two magnets to be equal and equal as per the formula.