Question

A bar magnet A of magnetic moment \[{{M}_{A}}\] is found to oscillate at a frequency twice that of magnet B of magnetic moment \[{{M}_{B}}\] when placed in a vibrating magneto-meter.
A. ${{M}_{A}}=2{{M}_{B}}$
B. ${{M}_{A}}=8{{M}_{B}}$
C. ${{M}_{A}}=4{{M}_{B}}$
D. ${{M}_{B}}=8{{M}_{A}}$

Answer 1

Hint: In this question, magnetic moment (M) and frequency (v) relationship is given by two bar magnets. Vibrating magnetometer is an instrument which is used to calculate the magnetic moments of a magnet placed in it by getting influenced by uniform magnetic field (B), the more the magnet aligns itself with external magnetic field, more magnetic moment and more oscillations will occur. As per question we taken two bar magnets which is placed in the vibrating magnetometer and the magnetic moment of second bar magnet is found to be double of magnetic moment of first magnet such as \[{{M}_{B}}\] is the magnetic moment of second bar magnet, \[{{M}_{A}}\] is the magnetic moment of the first bar magnet.

Formula used:
Using magnetic moment M, external magnetic field, B and moment of inertia, I, we can calculate time period of oscillation of magnet (time in which first oscillation is completed) such as,
\[T=2\pi \sqrt{\dfrac{I}{M{{B}_{H}}}}\]
More the magnetic moment less will be time to complete one oscillation.
The frequency is equal to reciprocal of time period such as,
\[\nu \text{ }=\text{ }1/T\]
\[\Rightarrow \nu =\dfrac{1}{2\pi \sqrt{\dfrac{I}{M{{B}_{H}}}}}\]
\[\Rightarrow \nu =2\pi \sqrt{\dfrac{M{{B}_{H}}}{I}}\]
Thus, more the magnetic moment (more alignment), more will be oscillation in one second which is termed as frequency.
$v\propto \sqrt{M}$

Complete step by step solution:
According to formula of frequency we get,
$v=\dfrac{1}{2\pi }\sqrt{\dfrac{M{{B}_{H}}}{I}} \\ $
$\Rightarrow v\propto \sqrt{M}$
Now, as the magnetic moment of the first bar magnet is \[{{M}_{A}}\] and the frequency will be \[{{v}_{1}}\]. Similarly as the magnetic moment of second bar magnet is \[{{M}_{B}}\] so, the frequency will be equal to \[{{v}_{2}}\] such as,
${{v}_{1}}\propto \sqrt{{{M}_{A}}}$and ${{v}_{2}}\propto \sqrt{{{M}_{B}}} \\ $
As given frequency (\[{{v}_{1}}\]) of first bar magnet is double of frequency (\[{{v}_{2}}\]) of second bar magnet such as
\[{{\nu }_{1}}=2{{\nu }_{2}}\]
\[\Rightarrow \sqrt{{{M}_{A}}}=2\sqrt{{{M}_{B}}}\]
Squaring on both side we get,
\[\therefore {{M}_{A}}=4{{M}_{B}}\]

Thus, the correct option is C.

Note:This question can also be simplified by dividing the frequency of first bar magnet to the frequency of seconds bar magnet such as,
$\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\ $
$\Rightarrow \dfrac{2}{1}=\sqrt{\dfrac{{{M}_{a}}}{{{M}_{b}}}}$
Squaring on both side we get,
$\Rightarrow 4=\dfrac{{{M}_{a}}}{{{M}_{b}}} \\ $
$\Rightarrow {{M}_{A}}=4{{M}_{B}}$
Near the earth's magnetic poles, a stunning effect caused by the magnetism of the planet can be seen. The northern aurora borealis and the southern aurora austral is are two names for this phenomenon. It is demonstrated via colored light patterns.