First slide

Magnetism

Question

A thin magnetic iron rod of length 30 cm is suspended in a uniform magnetic field. Its time period of oscillation is T second. It is broken into three equal parts perpendicular to axis. The time period of oscillation of one part when suspended in the same magnetic field is

Moderate

A

$T / \sqrt{3}$
B

$2 T / \sqrt{3}$
C

$\sqrt{3} T$
D

$4 T / \sqrt{3}$

Solution

$T = 2 π \sqrt{\frac{I}{M B}}$
Where given rod is broken into three equal parts then mass and length becomes one third but pole strength remains same.
We know that, $M a g n e t i c M o m e n t = P o l e S t r e n g t h \times l e n g t h; M o m e n t o f I n e r t i a = \frac{M a s s \times L e n g t h^{2}}{12}$
Hence, $M^{1} = \frac{M}{9}$ and $I^{1} = \frac{I}{27}$ (MI of each piece)
$T^{1} = 2 π \sqrt{\frac{I^{1}}{M^{1} B}}$
$T^{1} = 2 π \sqrt{\frac{9 I}{27 M B}} = (2 π \sqrt{\frac{I}{M B}}) \frac{1}{\sqrt{3}} = \frac{T}{\sqrt{3}}$

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