Question

A thin magnetic needle oscillates in a horizontal plane with a period $T$. It is broken into $n$ equal parts by cutting perpendicular to the length. Find the time period of each part.

Answer 1

$T=2\pi \sqrt{\frac{I}{MB}}$

I = moment of inertia of rod.

M=magnetic moment of rod.

case (1)

mass=m

magnetic moment =M

$T=2\pi \sqrt{\frac{\frac{m{L}^2}{12}}{MB}}$

$\therefore \frac{T^{\prime }}{T}=\frac{1}{n}$

Case (2)

mass $=\frac{m}{n}$

magnetic moment $=\frac{M}{n}$

$T=2\pi \sqrt{\frac{\left(\frac{m}{n}\right)\ast {\left(\frac{L}{n}\right)}^2\ast \frac{1}{12}}{\frac{M}{n}B}}$

$\therefore T^{\prime }=\frac{T}{n}$