Search any question & find its solution
Question:
Answered & Verified by Expert
A deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling of the deflection is $T$. When the magnet is removed, the period of oscillation of the needle is $T_0$ before settling to $0^{\circ}-0^{\circ}$. If the earth's induced magnetic field is $B_H$, the relation between $T$ and $T_0$ is
Options:
Solution:
2585 Upvotes
Verified Answer
The correct answer is:
$T^2=T_0^2 \cos \theta$
In deflection magnetometer, field due to magnet $F$ and horizontal component $B_H$ of earth's field are perpendicular to each other.
$\therefore$ Net field is $\sqrt{F^2+B_H^2}$
So the time period
$T=2 \pi \sqrt{\frac{1}{M \sqrt{F^2+B_H}}}$
When magnet is removed
$T_0=2 \pi \sqrt{\frac{1}{M B_H}}$
Also, $\quad \frac{F}{B_H}=\tan \theta$
Dividing Eqs. (i) by (ii), we get
$\begin{aligned} \frac{T}{T_0} & =\sqrt{\frac{B_H}{\sqrt{F^2+B_H^2}}} \\ & =\sqrt{\frac{B_H}{\sqrt{B_H^2+\tan ^2 \theta+B_H^2}}}=\sqrt{\frac{B_H}{B_H \sqrt{\sec ^2 \theta}}} \\ & =\sqrt{\cos \theta} \\ \Rightarrow \quad T^2 & =T_0^2 \cos \theta\end{aligned}$
$\therefore$ Net field is $\sqrt{F^2+B_H^2}$
So the time period
$T=2 \pi \sqrt{\frac{1}{M \sqrt{F^2+B_H}}}$
When magnet is removed
$T_0=2 \pi \sqrt{\frac{1}{M B_H}}$
Also, $\quad \frac{F}{B_H}=\tan \theta$
Dividing Eqs. (i) by (ii), we get
$\begin{aligned} \frac{T}{T_0} & =\sqrt{\frac{B_H}{\sqrt{F^2+B_H^2}}} \\ & =\sqrt{\frac{B_H}{\sqrt{B_H^2+\tan ^2 \theta+B_H^2}}}=\sqrt{\frac{B_H}{B_H \sqrt{\sec ^2 \theta}}} \\ & =\sqrt{\cos \theta} \\ \Rightarrow \quad T^2 & =T_0^2 \cos \theta\end{aligned}$
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.