
The period of oscillations of a magnet is $2\,sec$. When it is re-magnetised so that the pole strength is $4$ times its period will be :
A. $4\,sec$
B. $2\,sec$
C. $1\,sec$
D. $\dfrac{1}{2}\,sec$
Answer
161.4k+ views
Hint: In this given question, we will use the formula of the time period of an oscillating magnet and observe how ‘T’ relates to the horizontal magnet field. Then, after writing two time period expressions for the given data and analysing them, we will have a ratio that results from the magnetic field.
Formula used:
The formula of the time period of oscillation of the magnet is given by
$T=2\pi \sqrt{\dfrac{I}{MB}}$
Here, $T$ is the time period of oscillation of the bar magnet, $I$ is the moment of inertia of the bar magnet, $M$ is the magnetic moment and $B$ is the magnetic field intensity of a bar magnet.
Complete step by step solution:
In the question, we have given the period of oscillation of a magnet is $2\,\sec $and the pole strength after re-magnetised is $4$ times its period. 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 bar magnet is given by,
$T=2\pi \sqrt{\dfrac{I}{MB}}$
As we know that the frequency of an oscillation is directly proportional to the square root of the magnetic field, so we have, $T\,\alpha \dfrac{1}{\sqrt{M}}$ …….(i)
From the given information, we have the pole strength after re-magnetised is $4$times its period, so:
$M=4$
Now, substitute the value of $M$ in the equation (i), then we obtain:
$T\to \dfrac{1}{\sqrt{4}}$ times
$\Rightarrow T\to \dfrac{1}{2}$times
Therefore, the time period will be:
${{T}_{1}}=\dfrac{T}{2}=\dfrac{2}{2}=1\,\sec $
Thus, the correct option is C.
Note: If the angle of dip is specified in the query, then you should directly link it to the magnetic field's horizontal component. You may only relate the information provided to you for determining the magnetic field values by keeping in mind the expression of the time period of bar magnet oscillation. Vertical and horizontal components make up the two parts of the magnetic field.
Formula used:
The formula of the time period of oscillation of the magnet is given by
$T=2\pi \sqrt{\dfrac{I}{MB}}$
Here, $T$ is the time period of oscillation of the bar magnet, $I$ is the moment of inertia of the bar magnet, $M$ is the magnetic moment and $B$ is the magnetic field intensity of a bar magnet.
Complete step by step solution:
In the question, we have given the period of oscillation of a magnet is $2\,\sec $and the pole strength after re-magnetised is $4$ times its period. 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 bar magnet is given by,
$T=2\pi \sqrt{\dfrac{I}{MB}}$
As we know that the frequency of an oscillation is directly proportional to the square root of the magnetic field, so we have, $T\,\alpha \dfrac{1}{\sqrt{M}}$ …….(i)
From the given information, we have the pole strength after re-magnetised is $4$times its period, so:
$M=4$
Now, substitute the value of $M$ in the equation (i), then we obtain:
$T\to \dfrac{1}{\sqrt{4}}$ times
$\Rightarrow T\to \dfrac{1}{2}$times
Therefore, the time period will be:
${{T}_{1}}=\dfrac{T}{2}=\dfrac{2}{2}=1\,\sec $
Thus, the correct option is C.
Note: If the angle of dip is specified in the query, then you should directly link it to the magnetic field's horizontal component. You may only relate the information provided to you for determining the magnetic field values by keeping in mind the expression of the time period of bar magnet oscillation. Vertical and horizontal components make up the two parts of the magnetic field.
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