Question

A magnetic needle is pivoted through its centre of mass and is free to rotate in a plane containing uniform magnetic field $200 \times {10^{ - 4}}T$. When it is displaced slightly from the equilibrium, it makes 2 oscillations per second, If the moment of inertia of the needle about its axis of oscillation is $0.75 \times {10^{ - 5}}kg{m^2}$, the magnetic moment of the needle is
$\begin{align}
  &{\text{A}}{\text{. 6 }}J{T^{ - 1}} \\
& {\text{B}}{\text{. 0}}{\text{.6 }}J{T^{ - 1}} \\
  &{\text{C}}{\text{. 0}}{\text{.06 }}J{T^{ - 1}} \\
  &{\text{D}}{\text{. 0}}{\text{.006 }}J{T^{ - 1}} \\
\end{align} $

Answer 1

Hint: The magnetic needle acts like a bar magnet and we know the expression for the time period of oscillations in a bar magnet which involves magnetic moment, the moment of inertia of magnet, and the magnetic field. Using the available values, we can find out the value of the magnetic moment from this expression.

Formula used:
The time period of the oscillations of a bar magnet is given by the following expression:
\[T = 2\pi \sqrt {\dfrac{I}{{mB}}} \] …(i)
Here m represents the magnetic moment of the bar magnet which has been placed in a magnetic field B

Complete step-by-step solution:
We are given a magnetic needle which is pivoted through its centre of mass and is free to rotate in a plane containing uniform magnetic field $200 \times {10^{ - 4}}T$. Therefore,
$B = 200 \times {10^{ - 4}}T$
When the needle is slightly displaced from its position, it executes simple harmonic motion whose frequency is given as
$\nu = 2{s^{ - 1}}$
Therefore, the time period of oscillations is equal to the reciprocal of the above frequency which is given as
$T = 0.5s$
We are also given the value of the moment of inertia of the needle about its axis of oscillation which is
$I = 0.75 \times {10^{ - 5}}kg{m^2}$
We are required to find out the magnetic moment of the magnetic needle which can done by using expression (i) in the following way.
\[\begin{align}
 & T = 2\pi \sqrt {\dfrac{I}{{mB}}} \\
  & \Rightarrow m = 4{\pi ^2}\dfrac{I}{{{T^2}B}} \\
\end{align} \]
Now inserting the known values of time period T , the moment of inertia I and the magnetic field B in this expression, we get
\[\begin{align}
 & m = 4{\pi ^2}\dfrac{{0.75 \times {{10}^{ - 5}}}}{{{{\left( {0.5} \right)}^2} \times 200 \times {{10}^{ - 4}}}} \\
   &= 0.059J{T^{ - 1}} \simeq 0.06J{T^{ - 1}} \\
\end{align} \]
This is the required answer and hence the correct answer is option C.

Note: It should be noted that the magnetic field of the magnetic needle is interacting with the external magnetic field around it which is producing the motion in the needle. At the equilibrium, the net force is zero but when displaced from its position, it results in an oscillation about its center of mass.