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Magnetic Circuit Test 1
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Magnetic Circuit Test 1
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  • Question 1/10
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    Magnetic flux will be _________ if the surface area vector of a surface is perpendicular to the magnetic field.
    Solutions

    Magnetic flux:

    The magnetic flux is defined as the number of magnetic field lines passing through a closed surface

    Flux can be expressed as

    ϕ=BAcosθ

    ϕ  is the magnetic flux

    B is the magnetic flux density

    A is the area

    θ is the angle between the surface area vector of a surface and magnetic field

    Given that, θ = 90°

    ϕ=BAcos90=0

  • Question 2/10
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    When the separation between two charges is made four times, the force between them

    Solutions

    Coulomb’s law:

    The force between two point charges is along the line joining between them is directly proportional to the product of point charges and inversely proportional to the square of the distance between them:

    FQ1Q2R2

    Q1 & Q2 must be static or at rest.

    The distance (R) between two-point charges R is very large compared to the dimension of a charged body

    F=KQ1Q2R2

    Where K=14πεo=9×109m/F

    F=14πεoQ1Q2R2

    εo=14πfQ1Q2R2C2Nm2

    εo=8.854×1012Fm=10936πFm

    Coulomb force follows the superposition principle

    Application:

    When the separation (R) between two charges is made four times, then force between them is

    F1R2

    Force is decreases sixteen times

  • Question 3/10
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    The unit of Magmetomotive Foce (MMF) is
    Solutions

    Magnetomotive force (MMF):

    The current flowing in an electric circuit is due to the existence of electromotive force similarly magnetomotive force (MMF) is required to drive the magnetic flux in the magnetic circuit. The magnetic pressure, which sets up the magnetic flux in a magnetic circuit is called Magnetomotive Force.

    The strength of MMF is equal to the product of the current and no. of turns of the coil

    MMF can be expressed as

    MMF = N I = (Flux) x (Reluctance)

    The SI unit of MMF is ampere-turn (AT)

  • Question 4/10
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    Calculate the MMF required to produce a flux of 0.015 Wb across an air gap 2.5 mm long having an effective area of 200 cm2?
    Solutions

    Magnetomotive force (MMF):

    The current flowing in an electric circuit is due to the existence of electromotive force similarly magnetomotive force (MMF) is required to drive the magnetic flux in the magnetic circuit. The magnetic pressure, which sets up the magnetic flux in a magnetic circuit is called Magnetomotive Force.

    The strength of MMF is equal to the product of the current and no. of turns of the coil

    MMF can be expressed as

    MMF = N I = (Flux) x (Reluctance)

    The SI unit of MMF is ampere-turn (AT).

    Calculation:

    The Airgap length lg = 2.5 mm

    Effective area A = 200 cm2

    air gap sg=lgμoA=2.5×1034π×107×200×104 = 9.95 × 104

    MMF = Flux × reluctance = 0.015 × 9.95 × 104 = 1492.5 AT

  • Question 5/10
    1 / -0.25

    Determine the value of permeance (in H) of a coil, when the flux through the coil is 40 Wb and the value of produced mmf is 30 Amp-turns.
    Solutions

    MMF:

    The strength of MMF is equal to the product of the current and no. of turns of the coil

    MMF can be expressed as

    MMF = N I = (Flux) x (Reluctance)

    The SI unit of MMF is ampere-turn (AT)

    Calculation:

    Given that, MMF = 30 Amp-turns

    Magnetic flux = 40 Wb

    Reluctance = MMF / flux = 30 / 40 Amp-turns / Wb

    Permeance is the reciprocal of reluctance.

    Permeance = 40/30 = 1.33
  • Question 6/10
    1 / -0.25

    What is the relation between magnetic field intensity H and magnetic flux density?
    Solutions

    B-H relation:

    The relation between magnetic field intensity (H) and magnetic flux density (B) is

    B = μ H Wb/m2

    μ = The magnetic permeability

    μo = Absolute permeability

    μ= Relative permeability of the medium

  • Question 7/10
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    Magnetic field intensity has the dimension

    Solutions

    Magnetic field intensity: It is magnetic field strength represented in terms of a number of magnetic field lines (flux) passing through a unit area perpendicular to the field lines.

    H=MMFl=NIl

    Dimensional Formula of magnetic field intensity is IL-1
  • Question 8/10
    1 / -0.25

    Hysteresis loss is ________ proportional to the area under the hysteresis curve. Also, it is ________ proportional to the number of cycles of magnetization per second:
    Solutions

    The energy loss associated with hysteresis is proportional to the area of the hysteresis loop. As the area of the hysteresis loop for a specimen is found to be large, the hysteresis loss in this specimen is also large.

    Hysteresis loss occurring in a material is,

    Wh=η×Bm1.6×f×V

    Where η is hysteresis constant

    f is frequency or number of cycles per second

    Bm is magnetic flux density

    V is the volume of the core

    Hence it is directly proportional to the number of cycles of magnetization per second
  • Question 9/10
    1 / -0.25

    Curie temperature is the temperature above which
    Solutions

    Curie temperature:

    • For the ferromagnetic materials, below the Curie temperature, the atoms are aligned and parallel causing spontaneous magnetism
    • Above the Curie temperature, the material is paramagnetic as the atoms lose their ordered magnetic moments when the material undergoes a phase transition
  • Question 10/10
    1 / -0.25

    The relative permeability of a medium is equal to (with M = magnetization of the medium and H = magnetic field strength)
    Solutions

    The magnetic flux density is given by

    B = μ0 (H + M)

    Where, M = magnetization of the medium and H = magnetic field strength

    μr is the relative permeability of a medium

    ⇒ μ0μrH = μ0 (H + M)

    ⇒ μrH = H + M

    ⇒ (μr - 1) H = M

    (μr1)=MH

    μr=1+MH  

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