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Magnetism and Matter Test -16

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Magnetism and Matter Test -16
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Weekly Quiz Competition
  • Question 1
    1 / -0
    A magnet of magnetic moment M and length 2l is bent at its mid-point such that the angle of bending is 60o. Now, the magnetic moment is
  • Question 2
    1 / -0
    A magnetic needle is kept in a non-uniform magnetic field. It experiences
  • Question 3
    1 / -0
    The effective length of a magnet is 31.4 cm and its pole strength is 0.5 A-m. What will be the magnetic moment if it is bent in the form of a semicircle?
  • Question 4
    1 / -0
    Directions: The following question has four choices out of which only one is correct.

    A short bar magnet, placed with its axis at 30o with an external magnetic field of 0.16 T, experiences a torque of magnitude 0.032 J. The magnetic moment of the bar magnet ( in Am2) is 
  • Question 5
    1 / -0
    The magnetism of a magnet is due to which of the following?
  • Question 6
    1 / -0
    Directions: For the statements of Assertion and Reason, mark the correct answer.

    (a) If both assertion and reason are true and reason is the correct explanation of assertion.
    (b) If both assertion and reason are true, but reason is not the correct explanation of assertion.
    (c) If assertion is true, but reason is false.
    (d) If both assertion and reason are false.

    Assertion:     A disc-shaped magnet levitates above a superconducting material that has been cooled by liquid nitrogen.
    Reason:     Superconductors repel a magnet.
  • Question 7
    1 / -0
    Directions: The following question has four choices out of which only one is correct.

    The length of a magnet is larger than its width and height. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and all the three parts are then placed on each other with their like poles together. The time period of this combination will be
  • Question 8
    1 / -0
    Directions: The following question has four choices out of which only one is correct.

    A bar magnet of length 3 cm has points A and B along its axis at distances of 24 cm and 48 cm, respectively, on the opposite sides.



    Ratio of the magnitudes of the magnetic fields at these points will be 
  • Question 9
    1 / -0
    For a paramagnetic material, the dependence of magnetic susceptibility  on absolute temperature is given by the relation
  • Question 10
    1 / -0
    Materials suitable for making electromagnets should have
  • Question 11
    1 / -0
    Which of the following is the strongest diamagnetic substance? 
  • Question 12
    1 / -0
    A magnet of magnetic moment 20 CGS units is freely suspended in a uniform magnetic field of intensity 0.3 CGS units. The amount of work done in deflecting it by an angle of 30o (in CGS units) is
  • Question 13
    1 / -0
    Which of the following has the highest magnetic permeability?
  • Question 14
    1 / -0
    Magnetic susceptibility of a diamagnetic substance
  • Question 15
    1 / -0
    At a point on the axial line of a bar magnet of dipole moment M, the magnetic potential is V. What is the magnetic potential due to a bar magnet of dipole moment M/4 at the same point?
  • Question 16
    1 / -0
    A magnetic dipole in a constant magnetic field has ;
    Solution
    $$U= - \vec{\mu}. \vec{B}= \mu B cos\theta$$----------1
    $$\tau = \mu B sin\theta$$--------------2
    From 1 and 2, we can say that magnetic field has zero potential energy when the torque is maximum. Hence the answer and option C is correct
  • Question 17
    1 / -0
    Relative permittivity and permeability of a material are $$\epsilon_{\mathrm{r}}$$ and $$\mu_{\mathrm{r}}$$, respectively. Which of the following values of these quantifies are allowed for a diamagnetic material? 
    Solution
    For a diamagnetic material, $$\mu_r$$ and $$\epsilon_r$$ should have following bounds $$0<\mu_r <1$$ and for any material $$\epsilon_r >1$$
  • Question 18
    1 / -0
    The magnetic lines of force inside a bar magnet:
    Solution

  • Question 19
    1 / -0
    The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a ling solenoid with $$1000$$ turns/cm.. The current that should be passed in the solenoid to demagnetise the ferromagnet completely is:

    Solution
    $$N=1000 turns/cm $$

    $$H= \dfrac{NI}{l}$$

    $$100 \dfrac{A}{M} = \dfrac{1000 \times I}{10^{-2}}$$

    $$100 A = 10^5 \times I $$

    $$10^{-3}A=I$$

    $$I=1mA$$

  • Question 20
    1 / -0
    A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through $$60^0$$. The torque needed to maintain the needle in this position will be:
    Solution
    Work done to rotate the needle from $$0^{\circ}$$ to $$60^{\circ}=MB(\cos 0^{\circ}-\cos 60^{\circ})=W=\dfrac{1}{2}MB$$........(1)

    Torque required to keep needle at $$60^{\circ}$$ from the magnetic field $$\tau=MB\sin60^{\circ}=\dfrac{\sqrt{3}}{2}MB$$........(2)

    From equation 1 and 2
    $$\tau=\dfrac{\sqrt3}{2}\times 2W$$

    $$\tau=\sqrt3W$$
    Option A
  • Question 21
    1 / -0
    Curie temperature is the temperature above which
    Solution
    Curie temperature is the temperature at which certain materials lose their permanent magnetic properties, to be replaced by induced magnetism.
    Permanent magnetism is caused by the alignment of magnetic moments and induced magnetism is created when disordered magnetic moments are forced to align in an applied magnetic field.  Higher temperatures make magnets weaker, as spontaneous magnetism only occurs below the Curie temperature. 
  • Question 22
    1 / -0
    An example of a perfect diamagnet is a superconductor. This implies that when a superconductor is put in a magnetic field of intensity $$B$$, the magnetic field $$B_S$$ inside the superconductor will be such that :
    Solution
    A diamagnet has a tendency to repel the magnetic field lines passing through it.
    For a perfect diamagnet or a semiconductor, the field lines wont enter the superconductor at all 
    Hence, the magnetic field inside the superconductor will be 0
  • Question 23
    1 / -0
    Three identical bars A, B and C are made of different magnetic materials. When kept in a uniform magnetic field, the field lines around them look as given in the figure . Make the correspondence of these bars with their material being diamagnetic (D), ferromagnetic (F) and paramagnetic (P) .

    Solution
    Magnet  A  is  allowing  all  the  field  lines  passing  through  it , therefore  it  is  ferromagnetic. 
    Magnet  B  is  repelling  the  field  lines  therefore  it  is  diamagnet.
    Magnet  C  is  attracting  the  field  lines  therefore  it  is  paramagnet.
  • Question 24
    1 / -0
    Needles $$N_{1}, N_{2}$$ and $$N_{3}$$ are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will 
    Solution
    Ferromagnetic substances are strongly attracted by a magnet while, paramagnetic and diamaganetic substances are weakly attracted and repelled respectively.
  • Question 25
    1 / -0
    A planar loops of wire rotates in a uniform magnetic field. Initially, at $$t = 0$$, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $$10 s$$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :
    Solution
    Time period, $$T = 10 s$$
    Ref. image I

    Axis of rotation $$\rightarrow $$ y-axis.

    $$\vec{B} = B(-\hat{k})$$

    ref. image II
    at some time t, area vector makes angle '$$\theta$$' with $$\vec{B}$$.
    flux, $$\phi = \vec{B}. \vec{A}$$

    $$= BA \cos \theta$$

    emf, $$\varepsilon= \left|\dfrac{-d \phi}{dt} \right| = \left(BA \sin \theta \dfrac{d \theta}{dt}\right)$$

    $$= BA \omega \sin \theta$$

    When $$\theta = \dfrac{\pi}{2} , \varepsilon = \varepsilon_{max}$$

    So, $$\theta = n \pi + \dfrac{\pi}{2} $$   n = integes 

    When $$\theta = 0, \varepsilon = \varepsilon_{min} = 0$$

    So, $$\theta = n \pi$$, n = integes

    for $$\varepsilon  = \varepsilon_{max}, \theta = \omega t = n \pi + \dfrac{\pi}{2}$$

    $$\Rightarrow t = \dfrac{n \pi}{\omega} + \dfrac{\pi}{2 \omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} + \dfrac{\pi}{2 \times \left(\dfrac{2 \pi}{T} \right)}$$

    $$t = \left(\dfrac{n \pi}{\dfrac{2 \pi}{T}} \right) + \dfrac{\pi}{2 \left(\dfrac{2 \pi}{T} \right)}$$

    $$= \dfrac{n T}{2} + \dfrac{T}{4} = \dfrac{n \times 10}{2} + \dfrac{10}{4}$$
    $$= 5n + 2.5$$

    $$= 2.5 s, 5 + 2.5 , 5 \times 2 + 2.5 ...$$

    $$= 2.5 s, 7.5 s, 12.5 s....$$

    for $$\varepsilon = \varepsilon_{min}, \theta = \omega t = n \pi$$

    $$\Rightarrow t = \dfrac{n \pi}{\omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} = \dfrac{n T}{2}$$

    $$\Rightarrow  t = \dfrac{n \times 10}{2} = 5n$$

    $$= 0 s, 5s, 10s, ...$$

    So, possible option is option (C).

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