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Magnetic Materials & Magnets Test - 3

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Magnetic Materials & Magnets Test - 3
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  • Question 1
    1 / -0.25

    Which of the following statements about earth 's magnetism is correct

    Solution

    According to recent researches the magnetic field of earth is considered due to a large bar magnet situated in earth 's core.
    It is considered that the north pole of this large magnet is situated at the geographical south of earth and vice versa and as the magnetic field due to a bar magnet is from north pole to south pole of the maget thus the earth 's magnetic field is considered from geographical south to geographical north which are respectively north and south poles of the bar magnet.

  • Question 2
    1 / -0.25

    magnetic permeability of the substance  μ is

    Solution

    μ= μ0 μr . μis the permeability of medium, μ0 is permeability of free space, and μr is relative permeability of the medium.

  • Question 3
    1 / -0.25

    A toroid wound with 60.0 turns/m of wire carries a current of 5.00 A. The torus is iron, which has a magnetic permeability of  μm=5000 μ0  under the given conditions. H and B inside the iron are

    Solution

    B=B0 +Bm
    = μ0 nI+ μmnI
    =nI(μ+ μm)
    =60x5(5000 μ0 + μ0 )
    B=300(5001 μ0 )
    B=1.88T
    H=B/μm
    =300x5001 μ0 /5000 μ
    =300AT

  • Question 4
    1 / -0.25

    A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?

    Solution

    Work required to turn the dipole is given by
    W=MB[cos θi - cos θf ]
    Here θi is the initial angle made by a dipole with a magnetic field and  is the final angle made by a dipole with a magnetic field.
    magnetic moment is normal to the field direction.
    so, θi =0 and θf =90
    W = 1.5 ×0.22 [ cos0 °- cos90 °]
    W = 0.33J
     
    magnetic moment is opposite to the field direction.
    so, θi =0 and θf =180
    now, W = 1.5 ×0.22 [ cos0 °- cos180 °]
    = 0.33 [ 1 - (-1) ]
    = 0.66 J

  • Question 5
    1 / -0.25

    In the magnetic meridian of a certain place, the horizontal component of the earth ’s magnetic field is 0.26G and the dip angle is  60o . What is the magnetic field of the earth at this location

    Solution

    The earth 's magnetic field is Be ​and its horizontal and vertical components are He ​and Hv ​
    cos θ= He ​​ /Be
    ∴cos60o = (​0.26 ×10−4 ​ / Be )T
    ⇒Be ​=(​0.26 ×10−4 )/ (½)​=0.52 ×10−4 T=0.52G

  • Question 6
    1 / -0.25

    Which of the following statements about bar magnets is correct ?

    Solution

    Explanation:As magnetic monopole does not exist. If we split the bar magnet into two pieces  each part will have its own north and south pole.

  • Question 7
    1 / -0.25

    Magnetic material differences are explained by

    Solution

    All substances show some kind of magnetic behaviour. After all, they are made up of charged particles: electrons and protons. It is the way in which electron clouds arrange themselves in atoms and how groups of these atoms behave that determines the magnetic properties of the material. The atom (or group of atoms) in effect becomes a magnetic dipole or a mini bar magnet that can align according to the magnetic field applied. The net effect of all these dipoles determines the magnetic properties of the Magnetic Materials.

  • Question 8
    1 / -0.25

    A cube-shaped permanent magnet is made of a ferromagnetic material with a magnetization 500M of about The side length is 20 cm. Magnetic dipole moment of the magnet is.

    Solution

    Correct Answer :- C

    Explanation : Magnetic Moment = Magnetization ×Volume

    Side of cube = 20 cm

    = 0.2 m

    = 500 ×0.2 ×0.2 ×0.2

    = 4Am2

  • Question 9
    1 / -0.25

    A bar magnet of magnetic moment 1.5 J/T lies perpendicular to the direction of a uniform magnetic field of 0.22 T. What is the torque acting on it?

    Solution


     =1.5x0.22
     =15x22x10-3
     =330x10-3
    Τ=0.33J
     

  • Question 10
    1 / -0.25

    A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth ’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null –point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth ’s magnetic field.)

    Solution

    At null point (and along the axis), earth 's magnetic field and bar 's magnetic field are opposite in direction.
    On the equatorial line, the bar 's magnetic field is opposite in direction to its field on the axis. Hence, on the equatorial line, the two fields add up.
    As the null points are on the axis of the bar magnet, therefore,
    B1= (μ0/4 π)(2M/d3)=H
    On the equatorial line of magnet at same distance(d), field due to the magnet is  
    B2= (μ0/4 π)(M/d3)=B1/2=H/2
    Therefore, total magnetic field at this point on equatorial line is  
    B=B2+H=3H/2=3 ×0.36G/2=0.54G
     

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