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Magnetism on child

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data

    A child sleeps at an average distance of 30cm from household wiring. The mains supply is 240V r.m.s. Calculate the max possible magnetic flux density in the region of child when the wire is transmitting 3.6 kW of power.

    Thats the first part. I dont really understand what r.m.s is and how to convert it to normal V..


    2. Relevant equations

    VI=P
    F=BIL

    3. The attempt at a solution

    B= (P U0) / (V 2pi r)
    = (3.5 * 10^3)(4pi * 10^ -7) / (240 2pi (0.3))

    Getting 9.7 * 10 - 2 - not correct.
     
  2. jcsd
  3. Jan 2, 2010 #2

    mgb_phys

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    rms is just a way of stating the 'average' voltage

    Shouldn't flux have units of field/area?
     
  4. Jan 2, 2010 #3
    So how can u convert average voltage to actual voltage? on the working out they have a current of 21.2 A. How does that happen?

    are my formulas wrong?
     
  5. Jan 2, 2010 #4

    mgb_phys

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    It's AC so the 'actual voltage' changes continually - it has an average (in simple terms) voltage of 240V so an average current of 3600W/240V = 15A

    Now you need the equation for a field a certain distance from a current carrying wire.
     
  6. Jan 2, 2010 #5
    Ok i use

    B = (U0)(I)/ 2pi (0.3)
    = (4.7*10^-7)(14.58) / 2pi (0.3)
    = 3.64*10-6 T

    They have 21.2 A. and have B= 1.5*10-5 T
     
  7. Jan 3, 2010 #6

    tiny-tim

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    Welcome to PF!

    Hi NewAgeBerean! Welcome to PF! :wink:

    For AC voltage …

    Vrms = Vpeak/√2 = Vaverageπ/2√2 :smile:

    (same with AC current … Irms = Ipeak/√2)
     
  8. Jan 3, 2010 #7
    so in this case which V do i use - average or peak? Are my previous working out correct? and is the mark scheme answer to the question incorrect? (they used 21.2 A)
     
  9. Jan 3, 2010 #8
    The question asks for the maximum possible, not the average. The the peak value is about 1.414 (square root of 2) times the rms (root mean square) value.

    In real life the voltage in a circuit varies by plus or minus 10% from the nominal as well. Assuming the load's power remains constant, that could mean extra current if the voltage were low.
     
  10. Jan 4, 2010 #9

    tiny-tim

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    Looking at the question again, I'm confused by the use of "power" …
    … power varies with voltage in an AC current …

    P = V2/R, and so Ppeak = Vpeak2/R, Paverage = Vrms2/R.

    What exactly does the question mean? :confused:
     
  11. Jan 5, 2010 #10
    Its ok, i found out the answer they have is incorrect. Using P/V to get I gives me around 15A using the r.m.s. and around 10A using the peak value.

    They are using a current of 21.2A - where that comes from i dont know.

    Obviously they made a mistake.

    Thanks for all the replies
     
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