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B Difference in formulas of magnetic force and magnetic flux

  1. Mar 30, 2017 #1
    The magnetic force is defined as ## \rm F_e = Bqvsin\theta## and the magentic flux is defined as## \rm \phi = BAcos\theta##
    My question is since we are multiplying the two vectors, why is ##sin\theta## used in one equation and ##cos\theta## in second given that ##\theta## is the angle between the two vectors?
     
  2. jcsd
  3. Mar 30, 2017 #2

    berkeman

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    Staff: Mentor

    Can you say more about where you got those equations? What geometries are they for?
     
  4. Mar 30, 2017 #3
  5. Apr 3, 2017 #4

    PeterO

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    Homework Helper

    One is the force on a moving charge, one is the amount of flux threading a loop - not sure why you expect then to both have the same sin-cos function.
    The application where both "formulas" come together is when you have a loop spinning in a magnetic field.
    You can work out the expected current in the loop by considering the change in flux through the loop, or the force on each of the charges in the wire of the loop as it moves.
    (A square loop with (rather than a circular one) is easier to conceptualise using the moving charges method - but using the flux change method any shape loop will do.)
    According to the first formula, the maximum driving force on the charges occurs when they are moving perpendicular to the field - which occurs when the loop is "side-on" to the field and the electrons in the side wires are moving at right angles to the field as the loop rotates.
    In that situation, the flux threading the loop is zero.
    So the angle used wants a maximum for one "formula" and a minimum for the other. The sin and cos functions achieve that.
    (note: when the loop is side-on, the slightest rotation causes a (relatively) large change to the amount of flux that can thread the loop - so the "rate of change of flux" method will be returning maximum current induced in the loop at that time as well.
    When the loop is perpendicular to the field (square on), maximum flux threads the loop, but small rotations hardly change that amount, so little or no current is induced by rotation at that point. At that time, the electrons in the side wires are moving parallel to the field, so little to no force on them is created.
    Using degrees for ease of input to your calculator, find cos 0 - cos 1 (that's cos zero minus cos one) and compare it to cos 89 - cos 90. That will give you the relative sizes of change in flux for a 1 degree change in rotational position in the side-on, square-on positions.
     
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