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Homework Help: Induction Motors, Generators, Magnetic Field Strength, Force and Coupling

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Here are a bunch of questions that I've attempted. I feel fairly confident with my answers, but I just wanted to make sure :) Thanks!

    a) calculate the synchronous speed, amount of slip and the percentage slip in a typical aircraft a.c., four pole, 400hz induction motor whose speed on full load is 11400RPM

    b) calculate the frquency of an a.c. generator fitted witha six pole rotor and driven by a CSD rotating at an output shaft speed of 8000RPM

    ci)calculate the magnetic fiels strength of a solenoid coil of 1000 turns, suspended in air, carrying a current of 2 amps and concentrated over 10cm length.

    cii) Without changing the number of turns or current in the coil, how may the force exerted by the solenoid described in (ci) be increased by a considerably large factor?

    ciii) calculate the output voltage of a 10:1 step down transformer whose input is 240 volts and coupling efficiency is 95%

    2. Relevant equations
    Given in the attempt.

    3. The attempt at a solution
    a) synchronous speed=(120*frequency/poles) =>synchronous speed=12000RPM
    slip=1-(normal speed/synchronous speed) =>slip=0.05

    b) frequency=(poles*speed)/120
    =>frequency=(6*8000/120)=400hz. ---------
    I think you get one cycle for each pair of poles. If so that means 3 cycles per revolution.
    Frequency = (8,000rev/1min) X (3cycles/1rev) X (1min/60sec) = (400cycles/1sec) = 400 Hertz


    c)(i) Magnetic field at centre of solenoid is given by B =μₒNi /L
    [μₒ = 4π*10^-7 H/m (permeability of air), N = 1000 turns, i = 2A, L = 0.10m]

    B = (4π*10^-7)*1000*2 / 0.10 B = 0.025 T

    (ii) The field strength is increased dramatically by inserting an iron (or other ferromagnetic material) into the solenoid core (iron has a relative permeability of about 700, potentially able to increase the field strength by a factor of 700)

    (iii) At 100% efficiency the output voltage would be 24.0V (1/10 of input)
    At 95% the output is 0.95*24 = 22.80 V
  2. jcsd
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