With a standard rectangular bar magnet of length (l), breadth $(b ; b< < l)$ and magnetic moment $M$, the…
With a standard rectangular bar magnet of length (l), breadth $(b ; b< < l)$ and magnetic moment $M$, the time period of the magnet in a vibration magnetometer is $4 \mathrm{~s}$. If the magnet is cut normal to its length into four equal pieces, the time period (in seconds) with one of the pieces is
16
2
1
$1 / 4$
Solution
Time period of magnet in vibration magnetometer,
$T=2 \pi \sqrt{\frac{1}{M H}}$
where, $I=$ moment of inertia of magnet
$M=$ magnetic moment
$H=$ horizontal component of earth's magnetic field
$\begin{aligned}
& \therefore \quad \frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2} \cdot \frac{M_2}{M_1}} \\
& \Rightarrow \quad \frac{4}{T_2}=\sqrt{\frac{m l^2 / 12}{m / 4(l / 4)^2 / 12}} \\
& =\sqrt{\frac{4 \times 16 \times 12}{12 \times 4}}=\sqrt{16} \\
& \Rightarrow \quad \frac{4}{T_2}=4 \\
& \therefore \quad T_2=1 \mathrm{~s} \\
&
\end{aligned}$