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5.Magnetism and Matter
medium
$A $ magnet freely suspended in a vibration magnetometer makes $10 $ oscillations per minute at a place $A$ and $20 $ oscillations per minute at a place $B$. If the horizontal component of earth’s magnetic field at $A$ is $36 \times {10^{ - 6}}\,T$, then its value at $B$ is
A
$36 \times {10^{ - 6}}\,T$
B
$72 \times {10^{ - 6}}\,T$
C
$144 \times {10^{ - 6}}\,T$
D
$288 \times {10^{ - 6}}\,T$
Solution
(c)$\frac{{{T_A}}}{{{T_B}}} = \sqrt {\frac{{{{({B_H})}_B}}}{{{{({B_H})}_A}}}} \Rightarrow \frac{{60/10}}{{60/20}} = \sqrt {\frac{{{{({B_H})}_B}}}{{36 \times {{10}^{ – 6}}}}} $
$ \Rightarrow {({B_H})_B} = 144 \times {10^{ – 6}}\,T$
Standard 12
Physics
