Question

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

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

Answer 1

Answer: (B)

$T=2\pi \sqrt{\frac{I}{MB}}=2\pi \sqrt{\frac{w{l}^2/12}{\text{ Pole strength }\times 2l\times B}}$
$\therefore T\propto \sqrt{wl}$
$\therefore \frac{T_2}{T_1}=\sqrt{\frac{w_2}{w_1}\times \frac{l_2}{l_1}}$
$=\sqrt{\frac{w_1/2}{w_1}\times \frac{l_1/2}{l_1}}=\frac{1}{2}$
$\Rightarrow T_2=\frac{T_1}{2}=0.5s$