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Multiloop circuit, magnetic force on a current carrying wire, Ampere's Law

  1. May 30, 2007 #1
    1. Multiloop circuit
    In Fig. 27-71, R1 = 5.76 , R2 = 18.7 , and the ideal battery has emf = 13.8 V. (a) What is the size of current i1?


    Relevant equations
    i = V/R

    The attempt at a solution
    I redrew the 3 resistors (R2s) as parallel resistors. So i1 would be in the middle. Then all the resistances would have the same voltage 13.8V. So
    i1 = V/R2 = 13.8V/18.7 = 1.59 A. It's wrong though.

    2. Magnetic Force on a Current-Carrying Wire
    A wire 62.9 cm long carries a 0.480 A current in the positive direction of an x axis through a magnetic field with an x component of zero, a y component of 0.000230 T, and a z component of 0.0120 T. Find the magnitudes of x, y, and z components of the force on the wire.

    Relevant equations
    Magnitude of the force on a current carrying wire = (current)*(length)*(magnetic field)*(sin (angle between the directions of field and the length))

    The attempt at a solution
    The length of wire is along the x axis, so the angle between the wire and the x axis = 0. The angle between the x axis and y axis = 90 degrees, between x and z is also 90 degrees:
    force magnitude of x component = 0
    force magnitude of y component = 0.480A x 0.629m x 0.000230T x sin 90
    force magnitude of z component = 0.480A x 0.629m x 0.0120T x sin 90

    Only the first one's right.

    3. Ampere's Law
    In Fig. 29-63, a long circular pipe with outside radius R = 2.65 cm carries a (uniformly distributed) current i = 9.16 mA into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the magnitude of the current in the wire in milliamperes such that the ratio of the magnitude of the net magnetic field at point P to the magnitude of the net magnetic field at the center of the pipe is 3.93, but it has the opposite direction.


    Relevant equations
    Magnetic field at distance r outside long straight wire with current =
    (μ0 x i)/(2 x pi x r)
    Magnetic field around Amperian loop = μ0 x current enclosed

    The attempt at a solution
    (Magnetic Field at pt P) / (Magnetic Field at the center of the pipe) = 3.93

    so: (μ0 x i)/(2 x pi x R) = 3.93 x (μ0 x current enclosed)
    ...i = 3.93 x current enclosed x 2 x pi x R
    = 3.93 x 9.16 mA x 2pi x 0.0265m = 0.00599 (mA)(m)
    The units should only have mA so the answer is wrong.
  2. jcsd
  3. May 31, 2007 #2


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    Homework Helper

    1. You are corrrect in assuming that the R2s are in parallel and that they will share the same voltage. The emf divides over this parallel combination and R1 though. What you can assume is that the three parallel branches will draw the same current, I1. This means that 3 x I1 will flow through R1. Use the eq.

    Emf = V1 + V2

    to solve for I1.

    2. The components of the magnetic field and the length vector form a triad of vectors where the length vector points along the x-axis.

    The direction of the force components are obtained by rotating the length vector towards each of the magnetic field components.

    By rotating the length vector towards the y magnetic field component we obtain a force component in the negative z-axis direction. The z component of the force therefore involves the y component of the magnetic field.
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