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Magnetic Circuit Test 2
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Magnetic Circuit Test 2

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  • Question 1/15
    1 / -0.33
    Biot-Savart's law states the relation between magnetic intensity and
    Solutions

    According to Biot-Savart’s law, the magnetic field at a point due to the incremental element of length dl carrying a current is given by

    Where B is a magnetic field in Tesla

    I, is the current in the wire in Ampere

    r is the radius of the wire in meter

    From the above expression, it is clear that the magnetic field at a point due to the incremental element of length dl carrying a current is directly proportional to the current carried by the element.
  • Question 2/15
    1 / -0.33
    What will be value of magnetic field (in mT) at the centre of a 10 m long coil, when the coil has 500 turns and carrying a current of 10 A?
    Solutions

    Given that, length of a coil (l) = 10 m

    Number of turns (N) = 500

    Current (I) = 10 A

    Magnetic field, 
  • Question 3/15
    1 / -0.33
    Ampere’s circuital law can be applied ________ the conductor
    Solutions

    Ampere's circuital law:

    • It can be written as the line integral of the magnetic field surrounding a closed loop equals to the number of times the algebraic sum of currents passing through the loop.
    • Suppose a conductor carries a current I, then this current flow generates a Magnetic field that surrounds the wire.

    • According to this law, magnetic fields are related to the electric current produced in them.
    • The law specifies the magnetic field that is associated with a given current or vice-versa, provided that the electric field doesn’t change with time.
    • It can be applied outside the conductor only.
  • Question 4/15
    1 / -0.33
    The field at any point on the axis of a current carrying circular coil will be:
    Solutions
    A continuous electric current in a current carrying coil is divided into multiple electric current elements. Using the superposition principle and the Biot-Savart’s Law, each discrete element generates its own magnetic field and when it is integrated with each field that produces a resultant field and it is aligned parallel to the axis of the coil.
  • Question 5/15
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    What will be the magnitude of magnetic field (in T) action parallel to the rotation of the disc of diameter 30 cm when the magnitude of EMF induced between the axis of rotation and the rim of the disc is 15 V and the angular speed of rotation of disc is 25 revolution per second.
    Solutions

    Given that E = 15 V, d = 30 cm, N = 25 rps = 1500 rpm

    We know that, 

    ⇒ B = 8.48 T
  • Question 6/15
    1 / -0.33
    The force required to separate two surfaces with a contact area measuring 3 cm × 4 cm, when flux density normal to the surface is 0.6 tesla, will be
    Solutions

    Given that, Area = 3 × 10-2 × 4 × 10-2 = 12 × 10-4

    Flux density (B) = 0.6 T

    The force is given by, 

  • Question 7/15
    1 / -0.33
    Ampere’s circuit law and which of the following law in electrostatics are analogous?
    Solutions

    Ampere's circuital law:

    • It can be written as the line integral of the magnetic field surrounding a closed loop equals to the number of times the algebraic sum of currents passing through the loop
    • Suppose a conductor carries a current I, then this current flow generates a Magnetic field that surrounds the wire

    • According to this law, magnetic fields are related to the electric current produced in them
    • The law specifies the magnetic field that is associated with a given current or vice-versa, provided that the electric field doesn’t change with time
    • It can be applied outside the conductor only
    • It is analogous to Gauss law in electrostatics

    Gauss law: The total electric flux through any closed surface surrounding charges is equal to the amount of charge enclosed divided by the permittivity.

  • Question 8/15
    1 / -0.33
    Magnetic circuit with the cross-sectional area of 25 cm2 is to be operated at 50 Hz from 150 Vrms supply. The number of turns required to active a peak magnetic flux density of 2 T in the core flux is
    Solutions

    Given that, cross sectional area (A) = 25 cm2 = 25 × 10-4 m2

    Frequency (f) = 50 Hz

    Magnetic flux density (B) = 2 T

    Emf (E) = 150 V

    We know that,

    E = 4.44 fNBA 

    ⇒ 150 = 4.44 × 50 × N × 2 × 25 × 10-4

    ⇒ N = 135 turns
  • Question 9/15
    1 / -0.33
    Magnetic field intensity at centre of circular coil, of diameter 2 m and carrying a current of 4 A is
    Solutions

    Given that, diameter = 2 m

    Radius (R) = 1 m

    Current (I) = 4 A

    Magnetic field intensity at centre of a circular coil is given by

  • Question 10/15
    1 / -0.33
    What will be the magnitude of the induced EMF (in V) in a coil of area of 120 square centimetres with 250 turns, if the coil is removed from a magnetic field of 15 T action at right angles to the coil in 1 second.
    Solutions

    Given that, Area (A) = 120 square centimetres

    Number of turns (N) = 250

    Change in magnetic field (ΔB) = 15 T

    Change in time (Δt) = 1 second

  • Question 11/15
    1 / -0.33
    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

    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 = 1.3
  • Question 12/15
    1 / -0.33
    Two long parallel conductors carry 200 A. If the conductors are separated by 40 mm, the force per meter of length of each conductor will be
    Solutions

    Given that, I1 = I2 = 200 A

    d = 40 × 10-3

    We know that, force 

  • Question 13/15
    1 / -0.33
    A circular loop has its radius increasing at a rate of 1 m/s. The loop placed perpendicular to a constant magnetic field of 0.8 wb/m2. When the radius of the loop is 2 m, the emf induced in the will be
    Solutions

    Given that,

    Rate of radius (dr/dt) = 1 m/s

    Magnetic field (B) = 0.8 wb/m2

    Radius of the loop (r) = 2 m

    Emf induced 

    = 2π × 2 × 0.8 × 1 = 3.2π V

  • Question 14/15
    1 / -0.33
    A magnetic circuit requires 600 AT to produce a certain quantity of flux in a magnetic circuit. If its excitation coil has 120 turns and 8 Ohm resistance, the voltage to be applied in exciting coil is
    Solutions

    Given that, flux = 600 AT

    Number of turns (N) = 120

    Resistance (R) = 8 Ω

    Current (I) = 600/120 = 5 A

    Voltage applied (V) = IR = 5 × 8 = 40 V
  • Question 15/15
    1 / -0.33
    A point pole has a strength of 8π × 10-4 weber. The force on a point pole of 6π × 10-4 weber placed at a distance of 20 cm from it will be
    Solutions

    Pole strength (m1) = 8π × 10-4 weber

    Pole strength (m2) = 6π × 10-4 weber

    Distance (r) = 20 cm = 20 × 10-2 m

    Force is given by, 

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