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

$I = \dfrac{{m{l^2}}}{{12}}$, 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.