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Homework Help: Magnetic field on current loop, involves rotational energy

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A rectangular loop of sides a = 0.3 cm and b = 0.8 cm pivots without friction about a fixed axis (z-axis) that coincides with its left end (see figure). The net current in the loop is I = 3.2 A. A spatially uniform magnetic field with B = 0.005 T points in the +y-direction. The loop initially makes an angle q = 35° with respect to the x-z plane.
    https://tycho-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys122/autumn08/homework/09/rectangular_loop_MFR/p8.gif [Broken]

    The moment of inertia of the loop about its axis of rotation (left end) is J = 2.9 × 10-6 kg m2. If the loop is released from rest at q = 35°, calculate its angular velocity [tex]\omega[/tex] at q = 0°.

    2. Relevant equations

    this is kinetic energy:
    KE = 1/2I[tex]\omega^{2}[/tex] but moment of inertia is specified to be J, so

    KE = 1/2J[tex]\omega^{2}[/tex]

    this is potential energy for a loop with current and magnetic field acting on it:
    PE = [tex]\mu[/tex]Bcos[tex]\theta[/tex]
    moment vector [tex]\mu[/tex] = NIA where N = number of loops, I = current, A = area, B = magnetic field

    so, PE = NIABcos[tex]\theta[/tex]

    3. The attempt at a solution

    I recognize that the angular kinetic energy equation can be used to find the angular velocity. KE = 1/2J[tex]\omega^{2}[/tex].
    Final kinetic energy = initial potential energy. Initial potential energy = NIABcos[tex]\theta[/tex]. so, we could say

    1/2J[tex]\omega^{2}[/tex] = NIABcos[tex]\theta[/tex]

    the right side is potential energy. that equals

    N = 1 turn; I = 3.2 Amps; A = 0.003*0.008 = 2.5e-5 meters; B = 0.005 Teslas; [tex]\theta[/tex] = 35 degrees

    (1)(3.2)(2.5e-5)(0.005)cos(35) = 3.276608177 * 10^-7 Joules. we'll call this 3.2766e-7
    so, now we have

    1/2J[tex]\omega^{2}[/tex] = 3.2766e-7

    J = 2.9 × 10-6 kilogram meters squared. we'll call this 2.9e-6

    1/2(2.9e-6)[tex]\omega^{2}[/tex] = 3.2766e-7

    [tex]\omega[/tex] = [tex]\sqrt{(2*3.2766e-7)/2.9e-6}[/tex] = 4.753655581 * 10^-1

    BUT, the answer happens to be what I did except that they did not take the square root at the end.
    so, they got 2.259724138 * 10^-1.

    did i do something wrong? what i did makes sense in my mind. is their answer wrong? why did they not not take a square root at the end? is kinetic energy not = 1/2J[tex]\omega^{2}[/tex], but instead 1/2J[tex]\omega[/tex] without the square?

    i thank everybody in advance!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 17, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What's the final potential energy, when θ = 0?
  4. Dec 17, 2008 #3
    wow, now that you bring it up, i realize my mistake. you're right! i forgot the final potential energy. so, the answer works, but there is something that bothers me.

    Ki + Ui = Kf + Uf

    there is no initial kinetic energy because it came from rest, so that is zero

    Ui = Kf + Uf

    rearranging it,

    Ui - Uf = Kf

    NIABcos35 - NIABcos0 = 1/2J[tex]\omega^{2}[/tex]

    NIAB(cos35 - cos0) = 1/2J[tex]\omega^{2}[/tex]

    (1)(3.2)(2.5e-5)(0.005)(cos35 - cos0) = 1/2(2.9e-5)[tex]\omega^{2}[/tex]

    -7.233918228 = 1/2(2.9e-5)[tex]\omega^{2}[/tex]

    when i find the potential energy, initial PE - final PE, i get a negative number. so, when i set it equal to 1/2J[tex]\omega^{2}[/tex], i get a square root of a negative number, which is not possible. if i take the square root of the absolute value of that number, the answer is correct. is there something wrong with my signs?
  5. Dec 17, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes, your signs are messed up. The correct expression for PE is PE = -IABcosθ (note the minus sign).

    So Ui - Uf = (-IABcos35) - (-IABcos0) = IAB(cos0 - cos35).
  6. Dec 18, 2008 #5
    ahhhh, i see. well, that will solve my problem. thanks Doc Al!
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