Question

In an experiment with vibration magneto-meter the value of \[4{{\pi }^{2}}\dfrac{I}{{{T}^{2}}}\] for a short bar magnet is observed as \[36\times {{10}^{-4}}\]. In the experiment with deflection magnetometer with the same magnet the value of \[\dfrac{4\pi {{d}^{3}}}{2{{\mu }_{0}}}\] is observed as \[\dfrac{{{10}^{8}}}{36}\]. The magnetic moment of the magnet used is:
A) \[50A{{m}^{2}}\]
B) \[100A{{m}^{2}}\]
C) \[200A{{m}^{2}}\]
D) \[1000A{{m}^{2}}\]

Answer 1

Hint: Let us know about a vibration magnetometer in brief before proceeding with the solution. A vibration magnetometer is used for the comparison of magnetic moments and magnetic fields. This device works on the principle that whenever a freely suspended magnet in a uniform magnetic field is disturbed from its equilibrium position, it starts vibrating about the mean position. In short, we can say vibration magnetometers work on the principle of torque acting on the bar magnet.

Formula Used:
 \[m{{B}_{H}}=4{{\pi }^{2}}\dfrac{I}{{{T}^{2}}}\], \[\dfrac{m}{{{B}_{H}}}=\dfrac{4\pi {{d}^{3}}}{2{{\mu }_{0}}}\]

Complete step by step solution:
For a vibration magnetometer, we have \[m{{B}_{H}}=4{{\pi }^{2}}\dfrac{I}{{{T}^{2}}}\] where \[m\] is the magnetic moment of the magnet, \[{{B}_{H}}\] is the intensity of the horizontal component of the earth’s magnetic field and \[T\] is the period of oscillation or the vibration period.
From the equation of the working principle of a magnetometer, we have
\[\dfrac{m}{{{B}_{H}}}=\dfrac{4\pi {{d}^{3}}}{2{{\mu }_{0}}}\] where \[d\] is the distance of the magnet from the magnetometer and \[{{\mu }_{0}}\] is the permeability constant
Multiplying the two equations written above, we get \[{{m}^{2}}=4{{\pi }^{2}}\dfrac{I}{{{T}^{2}}}\times \dfrac{4\pi {{d}^{3}}}{2{{\mu }_{0}}}\]
The values of both the terms have been provided to us in the question. Substituting the values, \[4{{\pi }^{2}}\dfrac{I}{{{T}^{2}}}\]\[=36\times {{10}^{-4}}\], and \[\dfrac{4\pi {{d}^{3}}}{2{{\mu }_{0}}}\]\[=\dfrac{{{10}^{8}}}{36}\], we get,
\[\begin{align}
  & {{m}^{2}}=36\times {{10}^{-4}}\times \dfrac{{{10}^{8}}}{36} \\
 & \Rightarrow {{m}^{2}}={{10}^{4}} \\
 & \Rightarrow m=100A{{m}^{2}} \\
\end{align}\]

The magnetic moment of the magnet is \[100A{{m}^{2}}\] and hence option (B) is the correct answer.

Note: We have two independent equations that both relate the magnetic moment of the magnet with the horizontal component of the earth’s magnetic field. Most students make the mistake of using only the equation of the working principle of the magnetometer and transposing the terms of the equation to try and find the unknown quantity and end up complexing the problem. We obtained our solution through a simple product, so you have to keep your heads open.