First slide

Magnetism

Question

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

Moderate

A

$\frac{T}{4}$
B

$\frac{T}{16}$
C

$\frac{T}{32}$
D

$\frac{T}{2}$

Solution

Time period of oscillation of magnetic needle is $T = 2 π \sqrt{\frac{I}{M B}}$

$A f t e r c u t t i n g m a s s a n d m a g n e t i c d i p o l o e m o m e n t b e c o m e s \frac{1}{4} a n d l e n g t h a n d b r e a d t h b e c o m e s \frac{1}{2} I n i t i a l l y, I f o r a b a r m a g n e t i s I = \frac{m (l^{2} + b^{2})}{12}$
$F i n a l l y, I_{2} = \frac{m}{16} (\frac{l^{2} + b^{2}}{12}) = \frac{I_{1}}{16} I n i t i a l l y, T_{1} = T, I_{1} = I, M_{1} = M$
$F i n a l l y, T_{2} = ?, I_{2} = \frac{I}{16}, M_{2} = \frac{M}{4} W e k n o w, T i m e P e r i o d o f o s c i l l a t i o n = T = 2 π \sqrt{\frac{I}{M B}}$
$∴ \frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1}}{I_{2}}} \times \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow \frac{T}{T_{2}} = \sqrt{\frac{I \times 16}{I}} \times \sqrt{\frac{M}{4 M}} = 2 \Rightarrow T_{2} = \frac{T}{2}$

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