The period of oscillations of a magnetic need | vedclass

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Question

(B) The time period of a magnetic needle oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia

Vedclass · Accepted Answer

$0.5$

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(B) The time period of a magnetic needle oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field strength.For a needle of mass $m$ and length $l$,the moment of inertia is $I = \frac{ml^2}{12}$ and the magnetic moment is $M = q_m \cdot l$,where $q_m$ is the pole strength.Substituting these,$T = 2\pi \sqrt{\frac{ml^2/12}{q_m \cdot l \cdot B}} = 2\pi \sqrt{\frac{ml}{12 q_m B}}$.Since $T \propto \sqrt{ml}$,when the needle is cut in half,the new mass $m' = m/2$ and the new length $l' = l/2$. The pole strength $q_m$ remains the same.Thus,$T' = T \sqrt{\frac{m'l'}{ml}} = T \sqrt{\frac{(m/2)(l/2)}{ml}} = T \sqrt{\frac{1}{4}} = \frac{T}{2}$.Given $T = 1.0 \, s$,the new time period $T' = \frac{1.0}{2} = 0.5 \, s$.

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.....$s$.

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