Question

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 $4\, s$. If the magnet is cut normal to its length into four equal pieces, the time period (in seconds) with one of the pieces is

• A. 16
• B. 2
• C. 1
• D. 1 / 4

Answer 1

Answer: (C)

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
$\therefore \frac{T_{1}}{T_{2}} =\sqrt{\frac{I_{1}}{I_{2}} \cdot \frac{M_{2}}{M_{1}}} $
$\Rightarrow \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 \frac{4}{T_{2}} =4 $
$\therefore T_{2} =1\, s$