Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismProperties of magnetEasy2 minPYQ_2020

PhysicsEasysingle choice

A magnetic needle of negligible breadth and thickness compared to its length, oscillates in a horizontal plane with a period $T$ . The period of oscillation of each part obtained on breaking the magnet into $n$ equal parts and perpendicular to the length is

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Answer:

Solution:

$T = 2 π \sqrt{\frac{I}{M B}}$

on breaking n equal parts perpendicular to length

$I' = \frac{1}{12} (\frac{m}{n}) {(\frac{l}{n})}^{2}$ of each part

$I' = \frac{I}{n^{3}}, M' = \frac{M}{n}$

$\frac{T^{'}}{T} = \sqrt{\frac{I^{'}}{M^{'}} \times \frac{M}{I}} = \sqrt{\frac{1}{n^{3}} \times n} = \frac{1}{n} \Rightarrow T^{'} = \frac{T}{n}$

Subject:PhysicsTopic:MagnetismSubtopic:Properties of magnet

⏱ 2mℹ️ Source: PYQ_2020

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