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A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of the original magnet is ${T_0}$ in a uniform magnetic field $B$. Then, the time period of each part in the same magnetic field is
A. $\dfrac{{{T_0}}}{2}$
B. $\dfrac{{{T_0}}}{3}$
C. $\dfrac{{{T_0}}}{4}$
D. None of these

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Last updated date: 14th Apr 2024
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MVSAT 2024
Answer
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Hint First establish the magnetic moments and the moment of inertia of each of the magnet pieces cut from the original piece. Then substitute these values in the equation for the time period of a magnet.
Formula used
${T_0} = 2\pi \sqrt {\dfrac{I}{{MB}}} $ where $I$ is the moment of inertia of the magnet and $M$ is its magnetic moment, $B$is the magnetic field and ${T_0}$is the time period of the magnet.

Complete step by step answer
For the original magnet, the time period is given by the formula,
${T_0} = 2\pi \sqrt {\dfrac{I}{{MB}}} $ where $I$ is the moment of inertia of the magnet and $M$ is its magnetic moment.
Now when the magnet is cut into three equal parts, the magnetic moments of each of the parts become one-third of its original value $M$ such that $M' = \dfrac{M}{3}$ where $M'$is the magnetic moment of each of the parts.
Now moment of inertia is essentially defined as a quantity expressing a body’s tendency to resist angular acceleration.
Moment of inertia of the magnet about an axis perpendicular to its length through its one end is given as $\dfrac{{m{l^2}}}{3}$ where $m$is its mass and $l$is its length.
Now, when the magnet is cut into three equal parts, mass of each part becomes $\dfrac{m}{3}$ and length of each part becomes $\dfrac{l}{3}$
So, moment of inertia of each of such part is $I' = \dfrac{m}{3}\dfrac{{{{\left( {\dfrac{l}{3}} \right)}^2}}}{3} = \dfrac{1}{{27}}\dfrac{{m{l^2}}}{3} = \dfrac{I}{{27}}$
So, the new time period of each of the pieces is
$\begin{gathered}
  T = 2\pi \sqrt {\dfrac{{I'}}{{M'B}}} = 2\pi \sqrt {\dfrac{3}{{27}}\dfrac{I}{{MB}}} \\
   \Rightarrow T = \sqrt {\dfrac{1}{9}} 2\pi \sqrt {\dfrac{I}{{MB}}} = \dfrac{{{T_0}}}{3} \\
\end{gathered} $
Therefore, the time period of each of the magnetic pieces is $\dfrac{{{T_0}}}{3}$

So, the correct answer is B.

Note Magnetic lines of force are always closed as magnetic monopoles do not exist. Which means that a magnet will always have a north pole and a south pole no matter how many times it is cut into smaller pieces.