Magnetism Question 123
Question: 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
[EAMCET (Med.) 2001]
Options:
A) $ 36 \times 10^{-6} $ T
B) $ 72 \times 10^{6}$ T
C) $ 144 \times 10^{-6} $ T
D) $ 288 \times 10^{6} $ T
Show Answer
Answer:
Correct Answer: C
Solution:
$ \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 $
BETA
