Question

A small bar magnet $A$ oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^{\circ}$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be
A. $\dfrac{T}{\sqrt{2}}$
B. $T$
C. $\sqrt{2} T$
D. $2T$

Answer 1

Hint: When the needle is on a vertical plane perpendicular to the magnetic moment, use the time period formula. Use the same formula when the needle is perpendicular to the magnetic moment and lying horizontally. Keep in mind that the time frames are the same in both situations. The ratio of the magnetic field's vertical and horizontal components is equal to the tangent of the dip's angle.

Formula used:
In vertical plane, the time period of the needle is \[{T_1} = 2\pi \sqrt {\dfrac{I}{{M{B_V}}}} \]
In horizontal plane, the time period of the needle is \[{T_2} = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} \]
\[I = \]Moment of inertia
\[M = \]Magnetic moment
\[{B_V}\]- Horizontal components.
\[{B_H}\]- Vertical components.

Complete step by step solution:
The equation for the relationship between the magnetic moment and the time period of the needle in vertical plane is,
\[{T_1} = 2\pi \sqrt {\dfrac{I}{{M{B_V}}}} \]
Given the information that, a small bar magnet \[A\] oscillates in a horizontal plane with a period\[T\]at a place where the angle of dip is \[{60^\circ }\]. The relationship between the needle's time period in the vertical plane and the magnetic moment is,
\[t = 2\pi \sqrt {\dfrac{{\rm{I}}}{{{\rm{M}}{{\rm{B}}_{\rm{H}}}}}} \]

The relationship formula can also be written as,
\[{{\rm{t}}^\prime } = 2\pi \sqrt {\dfrac{{\rm{I}}}{{{\rm{MB}}}}} \]
Write the above expression in first derivative formula:
\[\dfrac{T}{{T'}} = \sqrt {\dfrac{{B'}}{B}} \\
\Rightarrow \dfrac{T}{{T'}} = \sqrt {\dfrac{B}{{{B_H}}}} \\ \]
Substitute the values and simplify, we obtain
\[\dfrac{T}{{{T^\prime }}} = \sqrt {\dfrac{1}{{\cos \phi }}} \\
\Rightarrow \dfrac{T}{{{T^\prime }}} = \sqrt {\dfrac{1}{{\cos 60}}} = \sqrt 2 \]
And the solution is,
\[\therefore {T^\prime } = \dfrac{T}{{\sqrt 2 }}\]

Hence, option A is correct.

Note: The intensity of the magnetic field B is a vector quantity. As a result, the intensity of the geomagnetic field can be divided into two components: vertical and horizontal. These two components are clearly located on the magnetic meridian. If the product of these two factors is R and the angle of dip is\[\theta \].