Two bar magnets having the same geometry with magnetic moments $m$ and \[2m\] are firstly placed in such a way that their similar poles are on the same side. Then, its time period of oscillation is \[{T_1}\] . Now, the polarity of one of the magnets is reversed. Then, the time period of oscillation is \[{T_2}\] , then:
A. \[{T_1} < {T_2}\]
B. \[{T_1} = {T_2}\]
C. \[{T_1} > {T_2}\]
D. \[{T_2} = \infty \]
Answer
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Hint: To answer this question we need to use the formula for the time period of the bar magnet. In the first case, we need to find the time period of the bar magnet when the polarity is not reversed. And then we need to find the time period of the bar magnet when the polarity is reversed. Then we need to compare both the values to find which the larger value is.
Complete step by step solution:
The formula for the time period of the bar magnet is given by,
\[T = 2\pi \sqrt {\dfrac{I}{{MH}}} \]
Here, \[T\]is the time period of the bar magnet
\[I\] is the moment of inertia
\[M\] is the magnetic moment
\[H\] is said to be the horizontal component of the magnetic field.
Given that two bar magnets have magnetic moments $m$ and \[2m\]. In the case when the polarity is not reversed that is when the bar magnets are placed in such a way that their similar poles are on the same side, the time period of the bar magnet will be,
\[{T_1} = 2\pi \sqrt {\dfrac{I}{{(2M + M)H}}} \]
\[{T_1} = 2\pi \sqrt {\dfrac{I}{{(3M)H}}} \] …….. (1)
In the case when the polarity is reversed that is when the bar magnets are placed in such a way that their similar poles are on the opposite side. The time period of the bar magnet will be,
\[{T_2} = 2\pi \sqrt {\dfrac{I}{{(2M - M)H}}} \]
\[{T_2} = 2\pi \sqrt {\dfrac{I}{{(M)H}}} \] …….. (2)
Therefore on comparing equations (1) and (2) we can conclude that,
\[{T_1} < {T_2}\]
Therefore the correct option is A.
Note:
The behaviour of the magnetic dipole in a uniform magnetic field is similar to the behaviour of the electric dipole in a uniform electric field. The magnetic dipole moment is a vector quantity and the direction of the magnetic moment is from the south pole of the magnet to the north pole of the magnet.
Complete step by step solution:
The formula for the time period of the bar magnet is given by,
\[T = 2\pi \sqrt {\dfrac{I}{{MH}}} \]
Here, \[T\]is the time period of the bar magnet
\[I\] is the moment of inertia
\[M\] is the magnetic moment
\[H\] is said to be the horizontal component of the magnetic field.
Given that two bar magnets have magnetic moments $m$ and \[2m\]. In the case when the polarity is not reversed that is when the bar magnets are placed in such a way that their similar poles are on the same side, the time period of the bar magnet will be,
\[{T_1} = 2\pi \sqrt {\dfrac{I}{{(2M + M)H}}} \]
\[{T_1} = 2\pi \sqrt {\dfrac{I}{{(3M)H}}} \] …….. (1)
In the case when the polarity is reversed that is when the bar magnets are placed in such a way that their similar poles are on the opposite side. The time period of the bar magnet will be,
\[{T_2} = 2\pi \sqrt {\dfrac{I}{{(2M - M)H}}} \]
\[{T_2} = 2\pi \sqrt {\dfrac{I}{{(M)H}}} \] …….. (2)
Therefore on comparing equations (1) and (2) we can conclude that,
\[{T_1} < {T_2}\]
Therefore the correct option is A.
Note:
The behaviour of the magnetic dipole in a uniform magnetic field is similar to the behaviour of the electric dipole in a uniform electric field. The magnetic dipole moment is a vector quantity and the direction of the magnetic moment is from the south pole of the magnet to the north pole of the magnet.
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