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Easy Ampere's Law Problem

  1. Oct 16, 2009 #1
    An infinitely long solid conducting wire of radius a = 2 cm centered on the z-axis carries a current I1 = 5 A out of the screen. The current is uniformly distributed over the cross-section of the wire. Co-axial with the wire is an infinitely long thin cylindrical conducting shell of radius of b = 5 cm that carries a current of I2 = 4 A into the screen.

    http://i662.photobucket.com/albums/uu347/TwinGemini14/Ampere.gif

    1) At x = 1 cm, y = 0 (within the central wire), the magnetic field points

    A) in the +x direction.
    B) in the +y direction.
    C) in the -y direction.

    :: This should be B, in the +y direction. Use the right hand rule. Current is going out of the page and curl fingers to show that at (1,0), it should point 'up'.
    ----------------
    2) In the region a < r < b (between the central wire and the thin shell), the magnitude of the magnetic field

    A) is constant.
    B) is porportional to r.
    C) is proportional to 1/r.

    :: I'm not sure how to explain it, but my gut feeling is that it is proportional to r. So B.
    -----------
    3) At x = 6 cm, y = 0 (outside of the shell), the magnetic field points

    A) in the -x direction.
    B) in the +y direction.
    C) in the -y direction.

    :: This is very similar to question 1. The answer should be C. Use the right hand rule and notice that at (6,0), the field points 'down'.
    -----------------
    4) What is the magnitude of the magnetic field at x = 1 cm, y = 0 (within the central wire)?

    A) 5 µT
    B) 10 µT
    C) 15 µT
    D) 20 µT
    E) 25 µT

    :: This one I cannot seem to figure out. HELP NEEDED HERE. I've tried finding the current density at a radius of x=1 and using ampere's law, but cannot seem to come to one of the 5 options. Any advice?

    B = (4pi*10^-7)(pi*(0.01^2)) / (2pi * 0.01) = 3.14*10^-8 WRONG!!!
    ------------------------

    5) What is the magnitude of the magnetic field at x = 6 cm, y = 0 (outside of the shell)?

    A) 0.78 µT
    B) 3.33 µT
    C) 4.51 µT
    D) 8.20 µT
    E) 12.7 µT

    :: B = ((4pi*10^-7)*(Ienclosed))/(2pi*r)
    B = (4pi*10^-7)*(5-4) / (2pi * 0.06)
    B = 3.33*10^-6

    ANSWER IS B.
    --------------------------------
    Can somebody please look over my answers? I cannot seem to figure out number 4. Thanks for the help in advance.
     
  2. jcsd
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