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Electromagnetic breaking

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=154354&d=1338984201.png

    Q1: explain how the setup shown in the diagram acts as a break?

    Q2: Electric and hybrid cars use regenerative breaking. At high speeds, the car can be slowed by reversing the direction of the electric motor so that it acts as a generator. Explain how this system operates in slowing down the car?
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    2. Relevant equations

    3. The attempt at a solution

    Answer 1: The aluminium plate is a good conductor of electricity. As the plate passes through the uniform field, the amount of flux linking with the plate changes and hence an emf is induced. Because of the nature of the aluminium plate (conductor) a small current will flow. The direction of this induced current is such as to oppose the change creating it.
    The current-carrying aluminium plate will produce its' own magnetic field that interacts with the external magnetic field. By flemmings LHR the force on the plate acts to the left and hence acts to decellerate the glider.

    Is this correct?

    Answer 2:
    I am not entirely sure how this works so would be grateful if someone could help me along with my explanation. I will give it a go, though

    From what I understand the induced current in the genetator opposes the current from the motor and acts as a break....I am very unsure though how to answer this question to gain the 3 marks assigned to it...

    Thanks for your help :)
     
  2. jcsd
  3. Jun 6, 2012 #2
    note: the currents induced are EDDY currents
     
  4. Jun 6, 2012 #3
    for Q2) i think there could be a switch connected to the coil that switches through a connection with a capacitor and another connection with the car battery . When the break is applied , the coil experiencing b field lines is connected to the capacitor , the induced emf (or electrical energy)by the generator is stored within the capacitor , and when one is to use the stored energy , the capacitor is discharged .

    I dont know much about capacitors , and therefore i don't know whether this works.
     
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