Electromagnetism problem

Discussion in 'Homework Help' started by tadm123, Apr 3, 2014.

  1. tadm123

    Thread Starter Member

    Nov 20, 2013
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    Sorry for the short title, it was the only way to get past the bug.

    Here the problem asks us to find the power delivered to R in a circuit, with an induced current coming from a power line.
    (Faraday's law: V=-\ointE*dl = I*R)

    [​IMG]

    Can anyone tell what to do first? How do I approach this problem?

    Thanks in advance.
     
    Last edited: Apr 3, 2014
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    513
    You are right to want to use Faraday's law of induction, but you need the area version.

    The induced EMF depends upon the area of the loop and the angle between the loop and the magnetic field.

    Since the loop is fixed the induced EMF varies from min to max as the angle changes.

    Is this enough for you to check up on Farady's Law?

    B = \int {B\cos \theta dA}

    E = \left( {\frac{{ - d\Phi }}{{dt}}} \right) = \left( { - \int {\frac{{d(B\cos \theta )}}{{dt}}dA} } \right)
     
  3. tadm123

    Thread Starter Member

    Nov 20, 2013
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    Would this be the correct answer?

    [​IMG]
     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Since the loop is of appreciable size perpendicular to the cable, don't you think the magnetic field will vary considerably within it? The farside of the loop is nearly three times the distance from the cable as the nearside. The extent of the loop paralle to the cable does not affect field strength.

    I see you were not put off by my mistake in my first formula.
    My apologies for that. It should of course read phi = integral Bcos(theta) dA.
     
  5. tadm123

    Thread Starter Member

    Nov 20, 2013
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    So should the distance R in the first formula to get the Magnetic field be = 20m+38m= 58m, instead of 20m?
     
  6. jjw

    Member

    Dec 24, 2013
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    You should calculate k*integral( dA/r ) from r0 to r1 where r0 is 20m and r1 is 58m.
    dA is w*dr where w is width of the loop.
    This has a solution which you probably know.
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    The emf generated depends on the total magnetic flux (phi webers) passing through the loop.

    But we know that the flux density (B webers per sq metre) is inversely proportional to the distance from the cable, so varies across the loop.

    However we can simplify the calculation so long as the loop has the correct orientation, which maximises the induction.

    The magnetic field around a straight conductor forms concentric rings perpendicular to the direction of the current.
    If the loop is oriented as shown in the diagrams the lines of flux will pass through perpendicularly.
    The loop may be anywhere around the cable as shown in fig2.

    For powerline frequencies, the wavelength is very large compared to 62 metres so we can say that at any instant the field around the cable passing through the loop is the same at any section along its 62 m length.

    So we only need to consider the variation of the field strength (B) with distance from the cable. If we can get an average we can multiply this by the loop area to obtain the total flux.

    This is a simple integral to obtain a figure for the average field strength \overline B

    I have given this in Fig1

    The total flux can then be substituted into to EMF equation.
     
  8. tadm123

    Thread Starter Member

    Nov 20, 2013
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    So B= 4PI*10^*(-7)*4000* ln(58/20) * 1/(58-20)
    B= 2.25 * 10^{-5}

    total flux= 62*38*2*10^{-5}
    total flux= 5.28*10^{-2}

    emf= flux/377= 1.4*10^{-4} V
     
    Last edited: Apr 6, 2014
  9. jjw

    Member

    Dec 24, 2013
    173
    31
    emf=flux*377 = 19.9 V
     
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