Question

A bar magnet has a magnetic moment equal to $5 \times 10^{-5} \mathrm{\Omega m}$. It is suspended in a magnetic field, which has a magnetic induction $B=8 \pi \times 10^{-4} \mathrm{~T}$. Then, the magnet vibrates with a period of vibration equal to $15 \mathrm{~s}$. Then the moment of inertia of magnetic well be :
A. $7.16 \times 10^{-7} \mathrm{~kg}-\mathrm{m}^{2}$
B. $14.32 \times 10^{7} \mathrm{~kg}-\mathrm{m}^{2}$
C. $3.58\times {{10}^{-7}}~\text{kg}-{{\text{m}}^{2}}$
D. None of these.

Answer 1

Hint: A magnetic moment is the quantity of magnetism possessed by a material. A bar magnet has a north and South Pole, so it possesses two magnetic moments (defined as the product of the magnitude of each individual Magnetic field surrounding the piece).

Complete step by step solution:
When a bar magnet is placed in an external magnetic field, it will attempt to align with the external magnetic field so that its poles align with the opposite poles of the external magnetic field. As a result, the bar magnet will oscillate around its equilibrium position for a long time before finally coming to rest. This oscillation's time period is calculated as,
\[{\rm{T}} = 2\pi \sqrt {\dfrac{1}{{{\rm{MB}}}}} \]

Given the information that a bar magnet has a magnetic moment equal to
\[{\rm{M}} = 5 \times {10^{ - 5}}\Omega - {\rm{m}}\]
The bar magnet is suspended in a magnetic field and has a magnetic induction
\[{\rm{B}} = 8\pi \times {10^{ - 4}}\;{\rm{T}}\]
Using formula of time period of magnet
\[{\rm{T}} = 2\pi \sqrt {\dfrac{I}{{{\rm{MB}}}}} \]
Where \[I = \]moment of inertia; \[m = \]magnetic moment; and \[B = \]magnetic field.

Substituting the values provides, we obtain
\[{{\rm{T}}^2} = 4{\pi ^2}\dfrac{I}{{{\rm{MB}}}}\]
Move all the variables to one side with constant on another side:
\[I = \dfrac{{{{\rm{T}}^2}}}{{4{\pi ^2}}}{\rm{MB}}\]
Substitute all the values to solve:
\[I= \dfrac{{{{(15)}^2}}}{{4 \times {{(3.14)}^2}}} \times 5 \times {10^{ - 5}} \times 8\pi \times {10^{ - 4}}\]
Simplify the above expression to make it less complicated form:
\[I= 7.16 \times {10^{ - 7}}\;{\rm{kg}} - {{\rm{m}}^2}\]
Therefore, the magnet vibrates with a period of vibration equal to\[15\;{\rm{s}}\]. Then the moment of inertia is \[7.16 \times {10^{ - 7}}\;{\rm{kg}} - {{\rm{m}}^2}\].

Hence, option A is correct.

Note: We've assumed that the bar magnet has no energy losses and can oscillate indefinitely. However, there are energy losses due to air resistance in the support that the bar magnet is attached to and suspended in the magnetic field. This phenomenon is used by explorers in magnetic compasses to determine the magnetic North pole.