
A thin rectangular magnet suspended freely has a period of oscillation of $4s.$ If it is broken into two halves each having half their initial length, then when suspended similarly, the time period of oscillation of each part will be:
A) $4s$
B) $2s$
C) $1s$
D) $4\sqrt{2}s$
Answer
134.4k+ views
Hint: When a rectangular magnet is broken into two halves each having half their initial length and the magnetic moment is also half
$L'=\dfrac{L}{2}$
$M'=\dfrac{M}{2}$
$M=$ Magnetic moment
Formula Used:
$T=2\pi \sqrt{\dfrac{I}{MB}}$
Complete step by step answer:
When a thin rectangular magnet is suspended freely in an external magnetic field, it experiences a force repulsion or attraction, due to which the thin rectangular magnet oscillates.
The formula for the time period of oscillation
$T=2\pi \sqrt{\dfrac{I}{MB}}$ ……………………….(i)
Where $I=$ Moment of inertia
$M=$ Magnetic moment
$B=$ Magnetic field
Moment of inertia
$I=\dfrac{M{{R}^{2}}}{12}$
$M=x\times l$ …………….(ii)
Where $x=$ pole strength
When the magnet is broken into two equal parts, then the new length of magnet is $\dfrac{l}{2}$
And new magnetic dipole moment
$M'=x\times \dfrac{l}{2}$ ……………..(iii)
From equation (ii)and (iii)
$M'=\dfrac{M}{2}$
And now moment of inertia
$\begin{align}
& \Rightarrow I'=\dfrac{M'{{\left( l' \right)}^{2}}}{12} \\
& \Rightarrow I'=\dfrac{\left( \dfrac{M}{2} \right)\times {{\left( \dfrac{l}{2} \right)}^{2}}}{12} \\
& \Rightarrow I'=\dfrac{M{{l}^{2}}}{8\times 12} \\
& \Rightarrow I'=\dfrac{I}{8} \\
\end{align}$
$T'=2\pi \sqrt{\dfrac{I'}{M'B}}$ ………………(iv)
From equation (iv) divided by equation (i)
$\Rightarrow \dfrac{T'}{T}=\dfrac{2\pi \sqrt{\dfrac{I'}{M'B}}}{2\pi \sqrt{\dfrac{I}{MB}}}$
$\Rightarrow \dfrac{T'}{T}=\sqrt{\dfrac{\dfrac{I'}{M'B}}{\dfrac{I}{MB}}}=\sqrt{\dfrac{I'}{M'}\times \dfrac{M}{I}}$
$\Rightarrow \dfrac{T'}{T}=\sqrt{\dfrac{\dfrac{I}{8}}{\dfrac{M}{2}}\times \dfrac{M}{I}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$
$\Rightarrow T'=\dfrac{T}{2}$
Given $T=4s$
$\begin{align}
& \Rightarrow T'=\dfrac{4}{2} \\
& \Rightarrow T'=2\sec \\
\end{align}$
So we can say that option B is correct, that is $2s$.
Note:In the formula, $B$ is the external magnetic field, not its own magnetic field of magnet and $B$ is the same for both the cases.
$L'=\dfrac{L}{2}$
$M'=\dfrac{M}{2}$
$M=$ Magnetic moment
Formula Used:
$T=2\pi \sqrt{\dfrac{I}{MB}}$
Complete step by step answer:
When a thin rectangular magnet is suspended freely in an external magnetic field, it experiences a force repulsion or attraction, due to which the thin rectangular magnet oscillates.
The formula for the time period of oscillation
$T=2\pi \sqrt{\dfrac{I}{MB}}$ ……………………….(i)
Where $I=$ Moment of inertia
$M=$ Magnetic moment
$B=$ Magnetic field
Moment of inertia
$I=\dfrac{M{{R}^{2}}}{12}$
$M=x\times l$ …………….(ii)
Where $x=$ pole strength
When the magnet is broken into two equal parts, then the new length of magnet is $\dfrac{l}{2}$
And new magnetic dipole moment
$M'=x\times \dfrac{l}{2}$ ……………..(iii)
From equation (ii)and (iii)
$M'=\dfrac{M}{2}$
And now moment of inertia
$\begin{align}
& \Rightarrow I'=\dfrac{M'{{\left( l' \right)}^{2}}}{12} \\
& \Rightarrow I'=\dfrac{\left( \dfrac{M}{2} \right)\times {{\left( \dfrac{l}{2} \right)}^{2}}}{12} \\
& \Rightarrow I'=\dfrac{M{{l}^{2}}}{8\times 12} \\
& \Rightarrow I'=\dfrac{I}{8} \\
\end{align}$
$T'=2\pi \sqrt{\dfrac{I'}{M'B}}$ ………………(iv)
From equation (iv) divided by equation (i)
$\Rightarrow \dfrac{T'}{T}=\dfrac{2\pi \sqrt{\dfrac{I'}{M'B}}}{2\pi \sqrt{\dfrac{I}{MB}}}$
$\Rightarrow \dfrac{T'}{T}=\sqrt{\dfrac{\dfrac{I'}{M'B}}{\dfrac{I}{MB}}}=\sqrt{\dfrac{I'}{M'}\times \dfrac{M}{I}}$
$\Rightarrow \dfrac{T'}{T}=\sqrt{\dfrac{\dfrac{I}{8}}{\dfrac{M}{2}}\times \dfrac{M}{I}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$
$\Rightarrow T'=\dfrac{T}{2}$
Given $T=4s$
$\begin{align}
& \Rightarrow T'=\dfrac{4}{2} \\
& \Rightarrow T'=2\sec \\
\end{align}$
So we can say that option B is correct, that is $2s$.
Note:In the formula, $B$ is the external magnetic field, not its own magnetic field of magnet and $B$ is the same for both the cases.
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