Question 125: Magnetism and Matter

Question

A thin rectangular magnet suspended freely has a period of oscillation equal to $$T.$$ Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $$T',$$ the ratio $${{T'} \over T}$$ is

Accepted Answer

KEY CONCEPT :
The time period of a rectangular magnet oscillating in earth's magnetic field is given by
$$T = 2\pi \sqrt {{I \over {\mu {B_H}}}} $$
where $$I=$$ Moment of inertia of the rectangular magnet
$$\mu  = $$ Magnetic moment
$${B_H} = $$ Horizontal component of the earth's magnetic field
Case 1 :
$$T = 2\pi \sqrt {{I \over {\mu {B_H}}}} $$ where $$I = {1 \over {12}}M{\ell ^2}$$
Case 2 :
Magnet is cut into two identical pieces such that each piece has half the original length. Then
$$T' = 2\pi \sqrt {{{I'} \over {\mu '{B_H}}}} $$
where $$I' = {1 \over {12}}\left( {{M \over 2}} ight){\left( {{\ell  \over 2}} ight)^2} = {I \over 8}$$ and $$\mu ' = {\mu  \over 2}$$
$$	herefore$$ $${{T'} \over T} = \sqrt {{{I'} \over {\mu '}} 	imes {\mu  \over I}} $$
$$ = \sqrt {{{I/8} \over {\mu /2}} 	imes {\mu  \over I}} $$
$$ = \sqrt {{1 \over 4}}  = {1 \over 2}$$

Answer

$${1 \over {2\sqrt 2 }}$$

Answer

$${1 \over 2}$$

Answer

$$2$$

Answer

$${1 \over 4}$$

Magnetism and Matter

A thin rectangular magnet suspended freely has a period of oscillation equal to $T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T', the ratio {{T'} \over T}$ is

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