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

Hint: 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.