Question

Magnets $A$ and $B$ are geometrically similar but the magnetic moment of $A$ is twice that of $B$. If $T_{1}$ and $T_{2}$ be the time periods of the oscillation when their like poles and unlike poles are kept together respectively, then $\dfrac{T_{1}}{T_{2}}$ will be
A. $\dfrac{1}{3}$
B. $\dfrac{1}{2}$
C. $\dfrac{1}{\sqrt{3}}$
D. $\sqrt{3}$

Answer 1

Hint: We must consider the quantity magnetic moment as a vector while executing the addition operation on them in order to solve this problem. We will get varied results depending on how these two bar magnets' magnetic moments are oriented in relation to one another.

Formula used :
We are aware that \[T = 2\pi \sqrt {\dfrac{{\rm{I}}}{{{\rm{mB}}}}} \] determines the duration of an oscillation of a magnet with magnetic moment m and moment of inertia in a field.
Where \[I = \]moment of inertia;\[m = \]magnetic moment; and \[B = \]magnetic field.

Complete step by step solution:
A magnetic material that is exposed to an external magnetic field experiences a moment of force because the North and South poles are subject to equal and opposing forces. The magnetic material is thus able to orient itself in the direction of the externally applied field thanks to the resulting moment of force.

When like poles are kept together,
\[m = mA + mB\]
When unlike poles are kept together,
\[m = mA - mB\]
The formula for like poles is kept together.
\[{{\rm{T}}_{{\rm{Sum}}}} = 2\pi \sqrt {\dfrac{{\left( {{{\rm{I}}_1} + {{\rm{I}}_2}} \right)}}{{\left( {{{\rm{M}}_1} + {{\rm{M}}_2}} \right){{\rm{B}}_{\rm{H}}}}}} \]

The formula for unlike poles is kept together.
\[{{\rm{T}}_{{\rm{diff }}}} = 2\pi \sqrt {\dfrac{{{{\rm{I}}_1} + {{\rm{I}}_2}}}{{\left( {{{\rm{M}}_1} - {{\rm{M}}_2}} \right){{\rm{B}}_{\rm{H}}}}}} \]

Divide the formula of like poles and unlike poles:
\[\Rightarrow \dfrac{{{{\rm{T}}_{\rm{s}}}}}{{{{\rm{T}}_{\rm{d}}}}} = \dfrac{{{{\rm{T}}_1}}}{{\;{{\rm{T}}_2}}} \\
\Rightarrow \dfrac{{{{\rm{T}}_1}}}{{\;{{\rm{T}}_2}}}= \sqrt {\dfrac{{{{\rm{M}}_1} - {{\rm{M}}_2}}}{{{{\rm{M}}_1} + {{\rm{M}}_2}}}}\\ \]
Substitute the values according to the formula:
\[ \Rightarrow \dfrac{{{{\rm{T}}_1}}}{{\;{{\rm{T}}_2}}} = \sqrt {\dfrac{{2{\rm{M}} - {\rm{M}}}}{{2{\rm{M}} + {\rm{M}}}}} \\ \]
Solve the above expression, we get
\[ \therefore \dfrac{{{{\rm{T}}_1}}}{{\;{{\rm{T}}_2}}}= \dfrac{1}{{\sqrt 3 }}\]
Therefore, the value of \[\dfrac{{{T_1}}}{{{T_2}}}\] will be \[\dfrac{1}{{\sqrt 3 }}\].

Hence, the option C is correct.

Note: The magnitudes in this problem were calculated directly based on the concept, but you can also use the straightforward formula for a vector's magnitude. We know that \[I = {I_A} + {I_B} = \]Sum of moment of Inertia of the two magnets.