Question

Two bar magnets of the same mass, length and breadth but having magnetic moments M and 2M are joined together pole to pole and suspended by a string. The time period of the assembly in a magnetic field of strength H is 3 seconds. If now, the polarity of one of the magnets is reversed and the combination is again made to oscillate in the same field, the time period of oscillation (in seconds) will be:
$\begin{align}
  & (A)6 \\
 & (B)3 \\
 & (C)3\sqrt{3} \\
 & (D)\sqrt{3} \\
\end{align}$

Answer 1

Hint: Since the two bar magnets are joined together, we cannot simply add their individual time periods. We need to calculate the time period of the combined resultant system. For this, we will consider the moment of inertia and the magnetic moment of the resultant system as the sum of the two moments of inertia and the sum of the two magnetic moments(if polarity is the same).

Complete answer:
Let the time period of the first system be given by the term ${{T}_{1}}$ .
And let the moment of inertia of the magnets be given by ${{I}_{1}}\text{ and }{{I}_{2}}$ .
Then, the time period of this combined system will be given by the following formula:
$\begin{align}
  & \Rightarrow {{T}_{1}}=2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{(2M+M)H}} \\
 & \Rightarrow {{T}_{1}}=2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{3MH}} \\
\end{align}$ [Let this expression be equation number (1)]
Now, in the second case the polarity of one of the magnets is reversed and it is made to oscillate again.
Let the time period of the new system be denoted by ${{T}_{2}}$ .
Then, the time period of this combined system will be given by the following formula:
$\begin{align}
  & \Rightarrow {{T}_{2}}=2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{[2M+(-M)]H}} \\
 & \Rightarrow {{T}_{2}}=2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{MH}} \\
\end{align}$ [Let this expression be equation number (2)]
On dividing equation number (2) by (1), we get:
$\begin{align}
  & \Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{MH}}}{2\pi \sqrt{\dfrac{{{I}_{1}}+{{I}_{2}}}{3MH}}} \\
 & \Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\dfrac{3}{1}} \\
\end{align}$
Given,
$\Rightarrow {{T}_{1}}=3s$
Putting this in the above equation, we get:
$\begin{align}
  & \Rightarrow {{T}_{2}}=\sqrt{\dfrac{3}{1}}\times 3s \\
 & \therefore {{T}_{2}}=3\sqrt{3}s \\
\end{align}$
Hence, the time period of the new system, when polarity of one of the magnets is reversed comes out to be $3\sqrt{3}s$ .

Hence, option (C) is the correct option.

Note:
On a rough note, it could be said that if in a combined system of two magnets, if one of the magnet’s poles is reversed keeping all the other parameters constant then the time period of the resulting system will increase. One could remember this as a property, as it may come in handy while solving problems like these.