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Rate of thermal energy

  1. Oct 20, 2005 #1
    Hi,

    My Question:

    A 22.0 cm- diameter coil consists of 20 turns of circular copper wire 2.6 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.65 x 10^-3 T/S.

    Determine the rate of which thermal energy is produced.

    My work:

    P= i^2 x R
    P= (.0013 A)^2(6.96 x 10^-4 ohms)
    P= 1.18 x 10^-19 W

    After this I'm lost. Should I look into trying to incorporate the rate, so that I can find the rate of which thermal energy is produced

    Thank You:smile:
     
    Last edited: Oct 20, 2005
  2. jcsd
  3. Oct 21, 2005 #2

    Tide

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    Where did your numbers for current and resistance come from?
     
  4. Oct 21, 2005 #3
    Got resistance by using

    R=rho(L)/(A), where rho=1.68 x 10^-8 ohm*m, L= 22 x 10^-2 m, and A= pi((2.6 x 10^-3 m)/2)^2

    so R= 6.96 x 10^-4 ohms

    And to find the current I first had to find the Emf which I used the following equation below

    Emf=-N(delta BA)/(delta t), where N=20 turns, and for the BA/t combo I used (-8.65 x 10^-3 T/s)(pi((2.6 x 10^-3 m)/2)^2)

    Emf=-(20 turns)((-8.65 x 10^-3 T/s)(pi((2.6 x 10^-3 m)/2)^2)
    )
    Emf=9.19 x 10^-7 volts


    Finally I used emf and R that I found to find the current

    I=(emf)/R
    I=(9.19 x 10^-7 volts)/(6.96 x 10^-4 ohms)
    I=.0013 A
     
  5. Oct 21, 2005 #4

    Tide

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    I didn't check your numbers but the argument looks good. The power you calculate IS the rate at which theremal energy is produced.
     
  6. Oct 21, 2005 #5

    lightgrav

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    Jena,

    the L in the resistance is the total length of the wire;
    in your case, 20 turns x 2 pi (.11m) total length.
    (so your R is too small by a factor 20 pi )

    The changing B-field is encircled by an Electric Field,
    where E 2 pi r = Delta(BA)/Delta(t) . Here,
    A is the Area that is pierced by the changing B-field,
    or the Area inside the encicling E-field loop (if smaller).

    If E is parallel (along the LENGTH of) your COIL of wire,
    the Voltage "accumulates" all along the wire, like
    Delta(V) = E Delta(s) = N 2 pi R_coil .
    So in this Delta V = N Delta(BA)/Delta(t) ,
    the Area extends outward to the COIL of wire
    (the place where the E-field makes a Voltage).
    (If the coil is bigger than the B-field region,
    (you only use the A where the B is going thru.)

    Looks like you used the cross-section Area of the wire,
    so your Area is too small by a factor of almost 10000.
     
  7. Oct 22, 2005 #6
    So to get the resistance I must first find the lenght

    L=(N)(2 pi(22*10^-2 m/2)) and use this in the equation to help me find my resistance.

    Is that what I'm supposed to do first.
     
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