Question

The period of oscillations of a magnetic needle in a magnetic field is 1.0 sec. If the length of the needle is halved by cutting it, the time perioid will be

• A. 2.0 sec
• B. 0.25 sec
• C. 0.5 sec
• D. 1.0 sec

Answer 1

The correct option is C 0.5 sec

Period of oscillation of magnetic needle is given as,

$\Rightarrow T=2\pi \sqrt{\frac{I}{M{B}_H}}$

$I$ = moment of inertia of oscillating magnet,
$M$= magnetic moment,
$B_H$ = external magnetic field

$T=2\pi \sqrt{\frac{km{l}^2}{Polestrength\times l\times B}}$

(k = any constant, m = mass of magnetic needle, l = magnetic length)

When a magnetic needle is cut half in length. the magnetic length of each part is half of original length but the pole strength remains same

$\therefore T\propto \sqrt{ml}$

$\Rightarrow \frac{T_2}{T_1}=\sqrt{\frac{m_2{l}_2}{m_1{l}_1}}$

$\Rightarrow \frac{T_2}{T_1}=\sqrt{\frac{\frac{m_1}{2}}{m_1}\times \frac{\frac{l}{2}}{l}}$

$\Rightarrow \frac{T_2}{1}=\frac{1}{2}$

$\therefore T_2=\frac{1}{2}$

$\Rightarrow T_2=0.5sec$

Hence, the correct option is (B)