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Magnetism and Matter

Section: Physics

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Question : 72 of 75

Marks: +1, -0

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is

2\ \text{s}

. The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be

$2 \sqrt{3}\ \text{s}$
$\frac{2}{3}\ \text{s}$
$2\ \text{s}$
$\frac{2}{\sqrt{3}}\ \text{S}$

Solution:

T = 2\pi \sqrt{\frac{1}{M \times B}}

where

I = \frac{1}{12} m \ell^2

When the magnet is cut into three pieces the pole strength will remain the same and

M. I.

(I') = \frac{1}{12} \left( \frac{m}{3} \right) \left( \frac{\ell}{3} \right)^2 \times 3 = \frac{I}{9}

We have, Magnetic moment

(M)

=

Pole strength

(m) \times \ell

\therefore

New magnetic moment,

=

Pole strength

(m) \times \ell

\therefore

New magnetic moment,

M' = m \times \left( \frac{\ell}{3} \right) \times 3 = m\ell = M

T' = \frac{T}{\sqrt{9}} = \frac{2}{3}\ \text{s}.

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