Question

A magnet is suspended horizontally in the earth’s field. The period of oscillation in the place is T. If a piece of wood of the same moment of inertia as the magnet is attached to it, the new period of oscillation would be:
A. \[\dfrac{T}{{\sqrt 2 }}\]
B. \[\dfrac{T}{2}\]
C. \[\dfrac{T}{3}\]
D. \[\sqrt 2 \,T\]

Answer 1

Hint: When the piece of wood of the same moment of inertia is attached to the magnet, its effective inertia will be doubled. Use the period of oscillation of the bar magnet placed in earth’s magnetic field for both the cases and divide the equations you obtained. The magnetic dipole moment does not change due to change in moment of inertia of the system.

Formula used:
\[T = 2\pi \sqrt {\dfrac{I}{{m{B_H}}}} \]
Here, I is the moment on inertia of the bar magnet, m is the magnetic dipole moment and \[{B_H}\] is the horizontal component of the earth’s magnetic field.

Complete step by step answer:
We have given that the bar magnet is suspended horizontally in the earth’s magnetic field. When it is displaced and then released, the period of its oscillations is T. We assume the magnetic dipole moment of this bar magnet is m and the horizontal component of earth’s magnetic field is \[{B_H}\].
We can express the time period of angular oscillations of the bar magnet as,
\[T = 2\pi \sqrt {\dfrac{I}{{m{B_H}}}} \] …… (1)
Here, I am the moment of inertia of the bar magnet.
Now, we have given that the piece of wood is attached to this bar magnet of the same moment of inertia about the axis of rotation. Therefore, the final moment of inertia of the system will be,
\[I' = 2I\]
We can express the time period of angular oscillations of the system of bar magnet and piece of wood as,
\[T' = 2\pi \sqrt {\dfrac{{2I}}{{m{B_H}}}} \] …… (2)
Here, the magnet dipole moment m will not change after attaching the piece of wood since it is non-magnetic material.
Dividing equation (2) by equation (1), we get,
\[\dfrac{{T'}}{T} = \dfrac{{2\pi \sqrt {\dfrac{{2I}}{{m{B_H}}}} }}{{2\pi \sqrt {\dfrac{I}{{m{B_H}}}} }}\]
\[ \Rightarrow \dfrac{{T'}}{T} = \dfrac{{\sqrt {2I} }}{{\sqrt I }}\]
\[ \therefore T' = \sqrt 2 T\]

So, the correct answer is option D.

Note: When you place a piece of wood in the earth’s magnetic field, it will not react to magnetic field lines of the earth’s magnetic field. Therefore, the magnetic dipole moment of the system is only due to the bar magnet suspended in the earth’s magnetic field. The moment of inertia of the bar magnet and piece of wood is the same and since it is attached to the bar magnet, the moment of inertia of the system will be the sum of the moment of inertia of the bar magnet and the piece of wood.