Question 184: Magnetism and Matter

Question

A small circular loop of conducting wire has
radius a and carries current I. It is placed in a
uniform magnetic field B perpendicular to its
plane such that when rotated slightly about its
diameter and released, it starts performing
simple harmonic motion of time period T. If the
mass of the loop is m then :

Accepted Answer

$$	au $$ = - MBsin $$	heta $$
I$$\alpha $$ = - MBsin $$	heta $$
for small $$	heta $$,
$$\alpha $$ = $$ - {{MB} \over I}	heta $$
$$ 	herefore $$ $${\omega ^2}$$ = $${{MB} \over I}$$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{I\left( {\pi {R^2}} ight)B} \over {{{m{R^2}} \over 2}}}} $$ = $$\sqrt {{{2I\pi B} \over m}} $$
$$ 	herefore $$ T = $${{2\pi } \over \omega }$$ = $$\sqrt {{{2\pi m} \over {IB}}} $$

Answer

$$T = \sqrt {{{2m} \over {IB}}} $$

Answer

$$T = \sqrt {{{\pi m} \over {IB}}} $$

Answer

$$T = \sqrt {{{\pi m} \over {2IB}}} $$

Answer

$$T = \sqrt {{{2\pi m} \over {IB}}} $$

Magnetism and Matter

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