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EEC 125 | Define the following elements as related to a magnetic circuit

Magnetic flux is the surface integral of the normal components of the magnetic field B over that surface. It is usually denoted as ∅. The SI unit of magnetic flux is Weber
∅ = BAcosø
∅ = Magnetic flux
B = Magnetic field
A = area
ø = Angle between a perpendicular vector to the magnetic field and magnetic flux
Magnetic flux density, B is the measure of the magnetic flux passing through a unit area in a plane at right angle to the flux. The SI unit of magnetic flux density is Weber per meter square (wb/m²), B = ∅/A
Magnemotive force (M.M.F) This is the flux which drives magnetic flux through a magnetic circuit (i.e the route or path which is followed by magnetic flux) and correspond to electromotive force (e.m.f) in an electric circuit. It’s usually measured in Ampere turns
F = IN, where F = Magnemotive force, I = current, N = number of turns

state three factors in which the force on a current carrying conductor in a magnetic field depends

I. Strength of magnetic field
II. Strength of electric current
III. Length of the conductor

Demonstrate the law of magnetic force of attraction and opposition with the aid of a diagram using a two bar magnets

The law of magnetic force of repulsion state that if two conductor carry current in opposite direction is said to be repulsion while the law of magnetic force of attraction state that if the two fluxes tend to neutralize each other in the space between conductor is said to be law of attraction i.e unlike pole attract each other.
  1. A conductor of active length 0.3m moves in a magnetic field at a linear velocity of 500m/s. If the magnetic flux density is 0.05T, calculate the average value of the induced e.m.f, if the direction of movement of the conductor is perpendicular to the field.
    The average value of the induced e.m.f = magnetic flux density × length × velocity
    Therefore e.m.f, E = BLV
    B = 0.05T, L = 0.3m, velocity= 500m/s
    0.3×0.05×500 = 7.5volts
  2. A coil of 1000 turns and length 0.2m carries a current of 5A. If the cross sectional area of the magnetic circuit is proportional to the flux 1.0cm² and the flux produced by the coil is 0.15mWb. Calculate the permeability of the magnetic material.
    L = 0.2m, N = 1000 turns, I = 5A, A = 1.0cm² = 0.01m², ∅ = 0.15mWb
    Length= L, number of turns= N, current= I, area = A, flux = 0.15×10–³Wb
    ∅ = BA; B = ∅/A; B = 0.00015/0.01
    Flux density, B = 0.015wb/m²
    Magnetic field=H
    H = IN/L
    (5 × 1000) ÷ 0.2 = 25,000A/m
    Calculate the permeability of the magnetic material
    Permeability of the magnetic material = flux density/ magnetic field
    B/H = 0.015/25000
    6 × 10–⁷ Tesla/Ampere meter

Oluwamuyide Peter

On the 4th of November I officially became a member of the exclusive 1st student with distinction after five years of no such record, in the history of The Polytechnic Ibadan, Faculty of Engineering to graduate with distinction as a DPP students since its establishment in 2011. My unrelenting power to solve problems, have made me to create a platform where student can get valid information anywhere, anyplace at anytime Evolving education world wide 🌎

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