Q. A magnet freely suspended in a vibration magnetometer makes 40 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) 9 $\times 10^{- 6}$T
C) 144 $\times 10^{- 6}$T
D) 228 $\times 10^{- 6}$T
Explanation
(b) Here, $n_{1} = 40,\ n_{2} = 20,\ H_{1} = 36 \times 10^{- 6}$T,
$H_{2} = ?$
$\frac{H_{2}}{H_{1}} = \frac{n_{2}^{2}}{n_{1}^{2}} = \frac{(20)^{2}}{(40)^{2}} = \frac{1}{4},$
$H_{2} = \frac{36 \times 10^{- 6}}{4} = 9 \times 10^{- 6}$T
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.