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

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

Magnetism and Matter

Solution:

$T=2 \pi \sqrt{\frac{I}{M B}} =2 \pi \sqrt{\frac{w l^{2} / 12}{\text { Pole strength } \times 2 l \times B}}$
$\therefore T \propto \sqrt{w l}$
$\therefore \frac{T_{2}}{T_{1}}=\sqrt{\frac{w_{2}}{w_{1}} \times \frac{l_{2}}{l_{1}}}$
$=\sqrt{\frac{w_{1} / 2}{w_{1}} \times \frac{l_{1} / 2}{l_{1}}}=\frac{1}{2}$
$\Rightarrow T_{2}=\frac{T_{1}}{2}=0.5\, s$