A long bar magnet of time period T is cut in to four equal parts by cutting it perpendicular to both length and breadth. The time period of each part is

Question

Infinity Learn · Accepted Answer

Time period of oscillation of magnetic needle $$isT=2$$$$π$$IMBAfter cutting mass and magnetic dipoloe moment becomes $$14$$ and length and breadth becomes $$12$$   Initially , I for a bar magnet is $$I=ml2+b212Finally$$ , $$I2=m16l2+b212=I116$$   Initially, $$T1=T$$ ,  $$I1=I$$  , $$M1=MFinally$$ , $$T2=$$? ,  $$I2=I16$$ ,  $$M2=M4$$ We know , Time Period of oscillation $$=$$ $$T=2$$$$π$$IMB∴ $$T1T2=I1I2$$×$$M2M1$$ ⇒$$TT2=I$$×$$16I$$×$$M4M=2$$ ⇒$$T2=T2$$

A long bar magnet of time period T is cut in to four equal parts by cutting it perpendicular to both length and breadth. The time period of each part is

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