
Using a bar magnet P, a vibration magnetometer has a time period of 2 seconds. When a bar Q (identical to P in mass and size) is placed on top of P, the time period is unchanged. Which of the following statements is true?
A. Q is of non-magnetic material.
B. Q is a bar magnet identical to P, and its north pole is placed on top of P 's north pole.
C. Q is of unmagnetized ferromagnetic material.
D. Nothing can be said about Q 's properties.
Answer
133.8k+ views
Hint:Use the direct formula for a bar magnet's time period in terms of the horizontal component of the earth's magnetic field to answer this question i.e.,$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $. The horizontal component will now be the resultant horizontal magnetic field when another magnet is placed on top of it, and it will combine with the horizontal component of the additional magnet.
Formula used:
The relation between time period of a bar magnet in terms of magnetic field 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 (horizontal component)
Complete step by step solution:
A bar magnet is a rectangular or quadrilateral piece of steel or iron that possesses permanent magnetic properties and has two poles, north and south. A solenoid is a type of electromagnet that can generate regulated magnetic fields by passing an electric current through it. It is a loop with a length greater than its diameter.
As we know that the relation between time period of a bar magnet in terms of magnetic field is given as,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
It is now considered that the time period is two seconds.
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} = 2 \\ $
Now when another magnet is kept on to it the resultant magnetic field gets added. The field will also become twice and the moment will also become twice of original value.
$\therefore {T^1} = 2\pi \sqrt {\dfrac{{2I}}{{M \times 2B}}} = T \\ $
This is only possible if Q is the same bar magnet as P and its north pole is positioned on top of P's north pole.
Hence the correct option will be B.
Note: The three factors that determine the strength and direction of the earth's magnetic field are as follows. the earth's magnetic field's horizontal component, magnetic inclination, or angle of dip, and magnetic declination. Near the magnetic poles, the magnetic field's intensity is highest, while it is at its lowest near the equator.
Formula used:
The relation between time period of a bar magnet in terms of magnetic field 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 (horizontal component)
Complete step by step solution:
A bar magnet is a rectangular or quadrilateral piece of steel or iron that possesses permanent magnetic properties and has two poles, north and south. A solenoid is a type of electromagnet that can generate regulated magnetic fields by passing an electric current through it. It is a loop with a length greater than its diameter.
As we know that the relation between time period of a bar magnet in terms of magnetic field is given as,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
It is now considered that the time period is two seconds.
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} = 2 \\ $
Now when another magnet is kept on to it the resultant magnetic field gets added. The field will also become twice and the moment will also become twice of original value.
$\therefore {T^1} = 2\pi \sqrt {\dfrac{{2I}}{{M \times 2B}}} = T \\ $
This is only possible if Q is the same bar magnet as P and its north pole is positioned on top of P's north pole.
Hence the correct option will be B.
Note: The three factors that determine the strength and direction of the earth's magnetic field are as follows. the earth's magnetic field's horizontal component, magnetic inclination, or angle of dip, and magnetic declination. Near the magnetic poles, the magnetic field's intensity is highest, while it is at its lowest near the equator.
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