Question

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $19\mathrm{\%}$. By doing this the periodic time of the magnetometer will:
A. Increase by $19\mathrm{\%}$
B. Decrease by $19\mathrm{\%}$
C. Increase by $11\mathrm{\%}$
D. Decrease by $21\mathrm{\%}$

Answer 1

Hint: A magnetometer is a device that measures the direction, strength, and change of a magnetic field at a specific location (on or near Earth, or in space). It primarily measures magnetic intensity and fields.

Formula used:
The time period of the magnetometer before heating is given by,
$T=2\pi \sqrt{{\displaystyle \frac{I}{MB}}}....(1)$
Here, $T$ is time period of oscillation of bar magnet, $I$ is moment of inertia of bar magnet, $M$ is magnetic moment and $B$ is magnetic field intensity of bar magnet.

Complete step by step solution:
In order for the model to be true, the net force exerted on the object at the pendulum's end must be commensurate to the displacement. As we know, the motion of a simple pendulum may be described by the basic harmonic motion and simple harmonic motion can also be used to describe molecular vibration.

The time period of the magnetometer before heating is given by,
$T=2\pi \sqrt{{\displaystyle \frac{I}{MB}}}....(1)$
In the question, we have the magnetic moment which is reduced by $19\mathrm{\%}$, then we have:
$M^{\prime }=M-19\mathrm{\%}M$
$\Rightarrow M^{\prime }=M-0.19M$
$\Rightarrow M^{\prime }=0.81M$

The time period of the magnetometer after heating is given by,
$T^{\prime }=2\pi \sqrt{{\displaystyle \frac{I}{M^{\prime }B}}}$
Now, substitute the obtained value of $M^{\prime }$in the above formula, then:
$T^{\prime }=2\pi \sqrt{{\displaystyle \frac{I}{0.81MB}}}$
 $\Rightarrow T^{\prime }={\displaystyle \frac{1}{0.9}}\times 2\pi \sqrt{{\displaystyle \frac{I}{MB}}}....(2)$

To determine the new time period for the magnetometer, subtract the equation $(1)$from $(2)$, then we obtain:
$\mathrm{\Delta }T=T^{\prime }-T$
$\Rightarrow \mathrm{\Delta }T={\displaystyle \frac{1}{0.9}}\times 2\pi \sqrt{{\displaystyle \frac{I}{MB}}}-2\pi \sqrt{{\displaystyle \frac{I}{MB}}}$
$\Rightarrow \mathrm{\Delta }T=2\pi \sqrt{{\displaystyle \frac{I}{MB}}}(0.1111)$
From the above equation, we notice that the value of $T$ is equal to $2\pi \sqrt{{\displaystyle \frac{I}{MB}}}$.
So,
$\mathrm{\Delta }T=T(0.1111)$
$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }T}{T}}=0.1111$
$\Rightarrow 11.11\mathrm{\%}\approx 11\mathrm{\%}$
Therefore, the new time period will increase by $11\mathrm{\%}$.

Thus, the correct option is C.

Note: Magnetometers are used for a variety of purposes and have applications in a variety of fields. They are used in detecting submarines, locating iron deposits in various geographical areas, and detecting metals deep within the earth. These days, the magnetometers are also used in electronic gadgets such as some smartphones to receive the messages from varying magnetic fields by other nearby electromagnets.