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Magnetic Field in a Rectangular Conducting Loop

  1. Apr 10, 2008 #1
    [SOLVED] Magnetic Field in a Rectangular Conducting Loop

    1. The problem statement, all variables and given/known data
    Figure 31-64a shows a rectangular conducting loop of resistance R = 0.010 , height H = 1.5 cm, and length D = 2.5 cm being pulled at constant speed v = 55 cm/s through two regions of uniform magnetic field. Figure 31-64b gives the current i induced in the loop as a function of the position x of the right side of the loop. For example, a current of 3.0 µA is induced clockwise as the loop enters region 1. What are the magnitudes and directions of the magnetic field in region 1 and region 2?


    2. Relevant equations
    [tex] EMF=BLV [/tex]

    [tex] i=EMF/R [/tex]

    3. The attempt at a solution
    Okay, I found the magnetic field in region 1 like so:
    3*10^-6A = EMF / 0.010 ohms
    EMF=3*10^-8 V

    3*10^-8 V = B * 0.015m * 0.55 m/s
    B = 3.64 microT for region 1

    However, for region 2 I must be making a mistake somewhere. Could someone point out my mistake please? Here's what I tried:

    -2*10^-6A = EMF/ 0.010 ohms
    EMF = 2*10^-8 V
    2*10^-8V = B * 0.55 m/s * 0.015m
    B=2.42 microT

    I have a feeling the mistake is in the length but I don't entirely understand how to interpret the problem statement. Thanks for your help!
    Last edited: Apr 10, 2008
  2. jcsd
  3. Apr 13, 2008 #2
    I tried taking the difference between magnetic fields 1 and 2 but it doesn't seem to be working. Can anyone offer some help?
  4. Apr 3, 2010 #3
    Re: [SOLVED] Magnetic Field in a Rectangular Conducting Loop

    I realize this post is over two years old, but for anybody who cannot figure out this problem and stumbles across this page, here is the answer for region 2.

    EMF(2) = i(1)R + i(2)R = R[ i(1) + i(2) ] = (0.010 ohm)[(3*10^-6A) + (-2*10^-6A)]
    EMF(2) = 1*10^-8V

    Then use the same equation that daimoku was using to find B(2) for region 2.

    B(2) = EMF(2) / (v * H) = (1*10^-8) / (0.55m/s * 0.015m) = 1.21 microT.

    Both B(1) and B(2) are out of the page.
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