Question

The time period of a magnet, oscillating in an external magnetic field, is T. If it is divided in four equal parts, along its axis and perpendicular to its axis (as shown in the diagram), then the time period of each part will be,

• A. $4T$
• B. $\frac{T}{4}$
• C. $\frac{T}{2}$
• D. $T$

Answer 1

The correct option is C $\frac{T}{2}$

The time period of a magnet oscillating in an external magnetic field is,
$T=2\pi \sqrt{\frac{I}{\mu B}}$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}\times \frac{{\mu }_2}{{\mu }_1}}$

Here, the magnet is being divided in 4 identical pieces, each of which will have length $\frac{l}{2}$ and magnetic pole strength $\frac{m}{2}$

so, ${\mu }_1=ml$
And, ${\mu }_2=\frac{m}{2}\times \frac{l}{2}=\frac{\mu }{4}$
$\therefore \frac{{\mu }_2}{{\mu }_1}=\frac{1}{4}$

$I_1=\frac{M_1{l}_1^2}{12}$

For each part, $M_2=\frac{M_1}{4}$ and $l_2=\frac{l_1}{2}$

$\Rightarrow I_2=\frac{M_2{l}_2^2}{12}=\frac{1}{12}\times \frac{M_1}{4}\times \frac{l_1^2}{4}$

$I_2=\frac{I_1}{16}\Rightarrow \frac{I_1}{I_2}=16$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{16\times \frac{1}{4}}$

$\therefore T_2=\frac{T}{2}$

Hence, $\left(C\right)$ is the correct answer.