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Induced current and voltage, magnetism

  1. Jun 23, 2008 #1
    1. The problem statement, all variables and given/known data

    a single square loop of high resistivity wire ( rho = 10^-6 ohm-meters) is placed in a constant magnetic field B of 0.3 Teslas and oriented so that the axis of rotation of the loop is perpendicular to B and in the plane of the loop. the loop rotates with an angular frequency of 300/second.

    a) if the square has a side length of 3 cm, what is the peak value of the voltage induced around the loop during the rotation?

    b) if the wire has a cross-sectional area of 10^-6 m^2, what is the value of the current induced in the loop?

    2. Relevant equations

    angular frequency, omega = 2pi(f) where f is frequency

    peak voltage, V_peak(sin(omega*t)) = L(dI/dt) where t is time, dt is change in time, dI is change in current

    voltage V = V_peak(omega*t + phi_v) where phi_v is phase angle = 90 deg

    period T = 2pi(m)/qB where m is mass, q is charge, B is magnetic field

    current I = I_peak(sin(omega*t + phi_i) where omega is angular freq., phi_i is phase angle = 90deg

    voltage induced, V = (A)dB/dt where dB is change in magnetic field, dt is change in time, A is area (3*3 = 9cm^2 )=

    3. The attempt at a solution

    i am sure i am missing equations, which equation involves magnetic field and the other givens? how does the resistivity come into play?

    angular freq, omega = 2pi(f)
    300 = 2pi(f)
    frequency, f = 300/2pi = 47.74 rot/sec

    using "V_peak(sin(omega*t)) = L(dI/dt)" it seems i need time, if i find time, i will be able to determine V_peak for part a, as omega is given, is dI assumed constant?

    i am stuck on a, so haven't attempted part b, help for either part appreciated.
     
  2. jcsd
  3. Jun 23, 2008 #2
    Find the component of the area normal to the magnetic field as a function of time.
     
  4. Jun 23, 2008 #3
    since for part a, the length of the square is given to be 3cm, than the area would be 9cm^2, or 0.09m^2.

    if i use " V = (A)dB/dt" ---> V = 0.09(0.3)/dt

    is this what you mean?

    how do i get time, is it related to the frequency i found in the original post?
     
  5. Jun 23, 2008 #4
    If I read your explanation correctly, isn't the square loop rotating so that the flux through the loop is changing?
     
  6. Jun 23, 2008 #5
    oh so magnetic flux = [integral(BdA)] where B is magnetic field, dA is change in area

    in part a, the area does not change, correct? so dA = 0, thus magnetic flux is zero, correct?

    you mention that the flux is changing, i'm confused, could you explain further? if flux is changing, i need to use faraday's law of induction then: emf, epsilon = -d(magnetic flux)/dt ---> where do i get epsilon, is this where i get the time portion for " V = (A)dB/dt" ---> V = 0.09(0.3)/dt??
     
  7. Jun 24, 2008 #6
    Is http://img128.imageshack.us/img128/3447/picturegs9.jpg a proper setup for this problem, or am I reading your explanation wrong?

    Assuming it is, as the wire rotates in that direction, the "face", if you will, of the loop of wire, goes from being completely perpendicular to the magnetic field to completely parallel to the magnetic field, so no, the change magnetic flux is not zero, it is some value that varies sinusoidally with the rotation of the loop

    Now, your magnetic field is constant, but the component of this loop's area that is perpendicular to the magnetic field at any given time is not. You use this "changing" area to get the change in flux which will, in turn, get you the induced voltage.
     
    Last edited: Jun 24, 2008
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