JEE Main 2003 — Magnetic Properties Of Matter Question with Solution

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by

$$T=2\pi \sqrt{\frac{I}{\mu B_H}}$$

where $$I=$$ Moment of inertia of the rectangular magnet

$$\mu =$$ Magnetic moment

$$B_H=$$ Horizontal component of the earth's magnetic field

Case 1 : $$T=2\pi \sqrt{\frac{I}{\mu B_H}}$$ where $$I=\frac{1}{12}M{\ell }^2$$

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

$$T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{{\mu }^{\prime }{B}_H}}$$

where $$I^{\prime }=\frac{1}{12}\left(\frac{M}{2}\right){\left(\frac{\ell }{2}\right)}^2=\frac{I}{8}$$ and $${\mu }^{\prime }=\frac{\mu }{2}$$

$$\therefore$$ $$\frac{T^{\prime }}{T}=\sqrt{\frac{I^{\prime }}{{\mu }^{\prime }}\times \frac{\mu }{I}}$$

$$=\sqrt{\frac{I/8}{\mu /2}\times \frac{\mu }{I}}$$

$$=\sqrt{\frac{1}{4}}=\frac{1}{2}$$