Question

A magnetic moment m oscillating freely in earth's horizontal magnetic field makes an oscillation per minute. If the magnetic moment is quadrupled and the earth's field doubled, the number of oscillation made per minute would becomes
A. $\dfrac{n}{2\sqrt{2}} \\ $
B. $\dfrac{n}{\sqrt{2}} \\ $
C. $2\sqrt{2n} \\ $
D. $\sqrt{2n}$

Answer 1

Hint:In the given question two cases are discussed with one magnet, in first case magnet is placed in the earth magnetic field (b), on once disturbance it started to align itself with the earth magnetic field (magnetic strength or magnetic moment) with magnetic strength (m) it started to oscillate and oscillation per minute is the frequency of oscillation (let say n). In second case we taken same magnet but earth magnetic field is made double as compared to first one (\[b'\text{ }=\text{ }2b\]) and thus magnetic moment four times as compared to first one (\[m'\text{ }=\text{ }4m\]), in that we need to find the frequency () or oscillation per minute.

Formula used:
The relationship between frequency and magnetic moment and field
\[ n=2\pi \sqrt{\dfrac{M{{B}_{H}}}{I}}\]
Where, $M$ is the magnetic moment, $B_H$ is the magnetic field and $I$ is the moment of inertia.

Complete step by step solution:
As per hint, we need to find the relationship mathematically between magnetic moment M, magnetic fields B and frequency V. As with increasing magnetic moment and magnetic field, the alignment increases and time to take one oscillation will decrease (time period decreases). Inversely relationship between time period (T) and magnetic moment and field such as,
\[T=2\pi \sqrt{\dfrac{I}{M{{B}_{H}}}}\]
($I$ is the moment of inertia but in this question we have taken the same magnet so it will be constant with $2\pi$.)
As frequency is define as the inverse of time period thus,
\[\text{n }=\text{ }1/T\]
So,
\[n=\dfrac{1}{2\pi \sqrt{\dfrac{I}{M{{B}_{H}}}}}\]
\[\Rightarrow n=2\pi \sqrt{\dfrac{M{{B}_{H}}}{I}}\]
So, there is a direct relationship between frequency and magnetic moment and field.

In the first case, the earth magnetic field is b and in that magnetic field the magnetic moment of the magnet is m. The number of oscillation made by this magnet in magnetic field in one minute or frequency of oscillation is n such as,
\[n=2\pi \sqrt{\dfrac{mb}{I}} \\ \]
\[\Rightarrow n=\sqrt{mb}\]

In second case, magnetic field of earth (b’) is get doubled (2b) as compared to magnetic field in first case and magnetic moment of magnet (m’) four times (4m) as compared to magnetic moment in first case. Then we need to determine the oscillation made by magnet in one second or frequency (n’) such as,
\[n'=2\pi \sqrt{\dfrac{m'b'}{I}} \\ \]
\[\Rightarrow n'=\sqrt{m'b'} \\ \]
As b’ = 2b and m’ = 4m then
\[n'=\sqrt{4m\times 2b} \\ \]
\[\Rightarrow n'=\sqrt{8\times mb} \\ \]
\[\Rightarrow n'=2\sqrt{2}\times \sqrt{mb}\]
Where root of mb is equal to frequency of oscillation of magnet in first case such as \[n=\sqrt{mb}\] then
\[\Rightarrow n'=2\sqrt{2}\times n\]

No. of oscillation per min = $\dfrac{1}{2\pi }\sqrt{\dfrac{M{{B}_{H}}}{I}}$
$n\propto \sqrt{M{{B}_{H}}} \\ $
$\Rightarrow M\to 4\,times \\ $
$\Rightarrow{{B}_{H}}=2\,times \\ $
$\Rightarrow v\to \sqrt{8\,times} \\ $
$\Rightarrow {{v}^{1}}=\sqrt{8}V \\ $
$\therefore v=2\sqrt{2}$

Thus, the correct option is C.

Note: As we first find the relation between time period and magnetic field and moment to find the frequency of oscillation of magnet in external earth magnetic field but it can be noticed that as we increase the magnetic field, oscillation of magnet will increases and thus, number of oscillation in one minute also increases (frequency increases). If we increase the magnetic moment of the magnet then, the magnet starts to align itself with the external magnetic field and thus, oscillation increases so, also frequency increases. Normally frequency is defined as oscillation per second but in this question it is measured in per minute.