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5.Magnetism and Matter
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A magnetic needle is made to vibrate in uniform field $H$, then its time period is $T$. If it vibrates in the field of intensity $4H, $ its time period will be
A
$2T$
B
$T/2$
C
$2/T$
D
$T$
Solution
(b)$T = 2\pi \sqrt {\frac{I}{{M{B_H}}}} \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{{({B_H})}_2}}}{{{{({B_H})}_1}}}} $
$ \Rightarrow {T_2} = T\sqrt {\frac{{{{(BH)}_1}}}{{{{(BH)}_2}}}} = \frac{T}{2}\;\;\;\left( {{{({B_H})}_2} = 4{{({B_H})}_1}} \right)$
Standard 12
Physics
