Time period of oscillation of a bar magnet in a uniform magnetic field is $T_{o}$. The magnet is cut into 3 equal parts transverse to its length. The new time period of oscillation of one part will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{T_o}{3}$
$T_o$, Initial time period of Magnet = $2π\sqrt{\frac{I}{MB}}$
where,
I = Moment of inertia
M = Magnetic moment of Bar Magnet
B = Magnetic field
When the Magnet is cut into 3 pieces,
$M'=\frac{M}{3}$ and $I'=\frac{I}{3^3}=\frac{I}{27}$
$T_f=2π\sqrt{\frac{\frac{I}{27}}{\frac{M}{3}×B}}=2π\sqrt{\frac{3}{27}\frac{I}{MB}}$
$=\frac{2πI}{MB}×\frac{1}{3}$
$=\frac{T_o}{3}$