Search any question & find its solution
Question:
Answered & Verified by Expert
A magnet freely suspended in a vibration magnetometer makes 40 oscillations per minute at place $A$ and 20 oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \mathrm{~T}$, then its value at $B$ is
Options:
Solution:
1528 Upvotes
Verified Answer
The correct answer is:
$9 \times 10^{-6} \mathrm{~T}$
Time period of magnet, $T=2 \pi \sqrt{\frac{I}{M H}}$
In both the condition magnet is same.
So,
$\begin{aligned} T & \propto \frac{1}{\sqrt{H}} \\ \frac{T_A}{T_s} & =\frac{\sqrt{H_n}}{H_A}\end{aligned}$
$\begin{gathered}\frac{60 / 40}{60 / 20}=\sqrt{\frac{H_n}{36 \times 10^{-4}}} \\ \frac{1}{2}=\frac{\sqrt{H_n}}{36 \times 10^{-6}}\end{gathered}$
Squaring both sides
$\frac{1}{4}=\frac{H_n}{36 \times 10^{-6}} \Rightarrow H_n=9 \times 10^{-4} \mathrm{~T}$
In both the condition magnet is same.
So,
$\begin{aligned} T & \propto \frac{1}{\sqrt{H}} \\ \frac{T_A}{T_s} & =\frac{\sqrt{H_n}}{H_A}\end{aligned}$
$\begin{gathered}\frac{60 / 40}{60 / 20}=\sqrt{\frac{H_n}{36 \times 10^{-4}}} \\ \frac{1}{2}=\frac{\sqrt{H_n}}{36 \times 10^{-6}}\end{gathered}$
Squaring both sides
$\frac{1}{4}=\frac{H_n}{36 \times 10^{-6}} \Rightarrow H_n=9 \times 10^{-4} \mathrm{~T}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.