Keeping dissimilar poles of two magnets of equal pole strength and length same side, their time period will be
A. One second
B. Zero
C. Infinity
D. Any values
Answer
Verified
116.4k+ views
Hint:We cannot just add the individual time periods of the two bar magnets since they are connected. The combined resultant system's time period must be determined. For this, we'll think of the system's moment of inertia and magnetic moment as being equal to the sum of its two moments of inertia and two moments of magnetic moment (if polarity is the same).
Formula used:
The time period of vibration magnetometer is,
$T = 2\pi \sqrt {\dfrac{I}{{mB}}} $
Where $I = $moment of inertia about the axis of rotation
$m = $Magnetic moment of magnet
$B = $Earth’s magnetic field
Complete step by step solution:
A bar magnet will rotate and attempt to align itself with the magnetic field when it is placed in a uniform magnetic field because of the force acting in the field's direction. Due to the rotational inertia caused by the torque, the magnet will rotate and attempt to align with the field, causing the bar magnet to move in a straightforward harmonic manner.
The torque acting on a bar magnet and rotational inertia are the basic operating principles of a vibration magnetometer. The time period of vibration magnetometer is given by,
$T = 2\pi \sqrt {\dfrac{I}{{mB}}} $
Keeping the opposite poles of two magnets with equivalent pole strengths and lengths. In this position we can say that,
$T = 2\pi \sqrt {\dfrac{{{I_1}}}{{({m_1} - {m_2})B}}} $
Since the pole strength is same, we can say that ${m_1} = {m_2}$
Putting the value, we get $T = \infty $
Hence option C is correct.
Note: On a rough note, it might be claimed that the time period of the resulting system will rise if one of the poles of a combined system of two magnets is flipped while maintaining all other parameters constant. We add the magnetic pole if the other poles remain similar.
Formula used:
The time period of vibration magnetometer is,
$T = 2\pi \sqrt {\dfrac{I}{{mB}}} $
Where $I = $moment of inertia about the axis of rotation
$m = $Magnetic moment of magnet
$B = $Earth’s magnetic field
Complete step by step solution:
A bar magnet will rotate and attempt to align itself with the magnetic field when it is placed in a uniform magnetic field because of the force acting in the field's direction. Due to the rotational inertia caused by the torque, the magnet will rotate and attempt to align with the field, causing the bar magnet to move in a straightforward harmonic manner.
The torque acting on a bar magnet and rotational inertia are the basic operating principles of a vibration magnetometer. The time period of vibration magnetometer is given by,
$T = 2\pi \sqrt {\dfrac{I}{{mB}}} $
Keeping the opposite poles of two magnets with equivalent pole strengths and lengths. In this position we can say that,
$T = 2\pi \sqrt {\dfrac{{{I_1}}}{{({m_1} - {m_2})B}}} $
Since the pole strength is same, we can say that ${m_1} = {m_2}$
Putting the value, we get $T = \infty $
Hence option C is correct.
Note: On a rough note, it might be claimed that the time period of the resulting system will rise if one of the poles of a combined system of two magnets is flipped while maintaining all other parameters constant. We add the magnetic pole if the other poles remain similar.
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