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Homework Help: Calculating the induced emf

  1. May 23, 2013 #1
    1. The problem statement, all variables and given/known data
    An aeroplane in horizontal flight has a wing span of 50 m and is travelling due north at a speed of
    500 kmph. Calculate the emf induced between the wing tips of the aeroplane.
    The value of the vertical component of the Earth’s magnetic field is 5.5 x 10-5 T.

    What additional information would be needed to determine whether the left wing tip was positive
    or negative?

    2. Relevant equations
    ε = BvL

    3. The attempt at a solution

    500 kmph = 500,000/3600 = 138.89 ms-1

    ε = 5.5 x 10-5 T x 50m x 138.89 ms-1
    ε = 0.38194V

    I hope thats right? lol

    Anyways I'm confused about the second part 'What additional information would be needed to determine whether the left wing tip was positive or negative?'
  2. jcsd
  3. May 23, 2013 #2
    do you know the right hand rule, used to determine the direction of any induced current (or emf)
  4. May 23, 2013 #3
    Yeah, but I don't see how this would help determine if the left wing tip was positive or negative :S
    Last edited: May 23, 2013
  5. May 23, 2013 #4


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    Staff: Mentor

    Consider a typical conduction electron in the metal of the plane; it's being carried through the magnetic field along with the plane. So, electron moving through magnetic field...
  6. May 23, 2013 #5
    you are given the vertical component of the Earth's field....Do you know its direction?
  7. May 23, 2013 #6

    rude man

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    Homework Helper
    Gold Member

    To determine emf direction, think of a unit positive charge along the wing. The force on it is F = q v x B so you need to know direction of v and B to determine the direction of F on the charge.

    The emf is simply the force on a unit charge times length L, or F*L which is work done on the unit charge. So for example if the force is from left wing to right wing then the left wing is - and the right wing is +. In vector notation, emf = ( v x B ) * L. So L defines the direction (and sign) of the emf. In the above example, L points to the right wing and so emf is +.
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