Question

A deflection magnetometer is adjusted in the usual way .when a magnet is introduced , the deflection observed is $\theta$, and the period of oscillation of needle in the magnetometer is T. When the magnet is removed , the period of oscillation is ${T_o}$. The relation b/w $T$and ${T_o}$ is:
A) ${T^2} = {T_o}^2\cos \theta$
B) ${T^2} = {T_o}\cos \theta$
C) $T = \dfrac{{{T_o}}}{{\cos \theta }}$
D) ${T^2} = \dfrac{{{T_o}^2}}{{\cos \theta }}$

Answer 1

We will first understand that there will be two types of magnetics field $ie,$ magnetic field due to magnet and magnetic field due to the earth . then we will find out the resultant magnetic field and then solve it further by using the formula for time period . Refer to the solution below.
Formula for time period is
$\Rightarrow T = 2\pi \sqrt {\dfrac{I}{{MB}}}$
Where I is current , M is magnetic moment and B is magnetic field.

Complete step by step answer:
In the usual setting of deflection magnetometer, the field due to magnet $\left( B \right)$ and horizontal component $\left( {{B_H}} \right)$ of earth ‘s field are perpendicular to each other . Therefore the net field on the magnetic needle is $\sqrt {{B^2} + {B_H}^2}$
By tangent law ,$B = {B_H}\tan \theta$
${B_{net}} = \sqrt {{B^2} + {B_H}^2}$
$\Rightarrow {B_{net}} = \sqrt {{B^2} + {B_H}^2}$
$\Rightarrow {B_{net}} = \sqrt {{B_H}^2 + {B_H}^2{{\tan }^2}\theta }$
Taking BH2 common
${B_{net}} = \sqrt {{B_H}^2(1 + {{\tan }^2}\theta )}$
Since
${\sec ^2}\theta = 1 + {\tan ^2}\theta$

\Rightarrow {B_{net}} = {B_H}\sqrt {{{\sec }^2}\theta } \\\ \Rightarrow {B_{net}} = {B_H}\sec \theta \\\ \Rightarrow {B_{net}} = B\dfrac{1}{{\cos \theta }} \\\ \ $$ Now the time period with magnet is given by $$T = 2\pi \sqrt {\dfrac{I}{{M{B_{net}}}}} $$ $$ \to $$$$\left( 1 \right)$$ Again time period when magnet is removed , $${T_0} = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} $$ $$ \to $$$$\left( 2 \right)$$ Dividing $$\left( 1 \right)$$by $$\left( 2 \right)$$ , we get $$\dfrac{T}{{{T_0}}} = \dfrac{{2\pi \sqrt {\dfrac{I}{{M{B_{net}}}}} }}{{2\pi \sqrt {\dfrac{I}{{M{B_H}}}} }}$$ $$ \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\dfrac{I}{{M{B_{net}}}} \times \dfrac{{M{B_H}}}{I}} $$ $$ \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\dfrac{{{B_H}}}{{{B_{net}}}}} $$ $$ \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\cos \theta } $$ $$\left[ \begin{gathered} {B_{net}} = \dfrac{{{B_H}}}{{\cos \theta }} \\\ \cos \theta = \dfrac{{{B_H}}}{{{B_{net}}}} \\\ \end{gathered} \right]$$ Squaring both sides $${\dfrac{T}{{{T_0}^2}}^2} = \cos \theta $$ $$ \Rightarrow {T^2} = {T_0}^2\cos \theta $$ Hence the relation between $$\operatorname{T} $$ and $${T_O}$$ is $${T^2} = {T_0}^2\cos \theta $$ **Therefore the answers is option (A).** **Note:** Magnetic field a neighbourhood vector field of a magnet , electric current or electric field changing , where magnetic fields like earth cause magnetic compasses and other permanent magnets to line up in the field direction.