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I Faraday's law or Lorenzt force?

  1. Feb 13, 2017 #1
    A student in my physics class posted, in a group, a wrong answer to a question.

    The situation was: A plane has a wire extended between the tips of its wings and flies through a magnetic field, perpendicularly, while accelerating.

    The question was: What will be the induced current?

    His answer was 0, and he explained that as the plane was flying through a uniform field, there would be no change in Magnetic flux density, so no change in flux linkage and ultimately no induced current.

    I stated, after clarifying with a question on here earlier today, that instead of consulting Faraday's laws, he should think about Fleming's left hand rule.

    My teacher told me I was wrong, and said as the plane is accelerating, it cuts the field lines at a greater rate, so according to Faraday's law, a voltage will be induced, but he also says this voltage will be increasing.... Even if it was Faraday's laws, as this acceleration was constant, wouldn't the voltage be constant?

    My next response is a lengthy one:

    Faraday's law dictates proportionality between an induced EMF and rate of change of flux linkage. The rule of a wire cutting through fields lines is contrarily a result of Fleming's left hand rule, where a magnetic force is exerted on the electrons inside the wire that is cutting the field, in the direction of the wire; this leads to a potential difference between both sides of a wire, or induced current for a closed circuit. Also, flux linkage, and thus Faraday's law, refer to solenoids, don't they? Ultimately, the induced voltage will indeed increase due to the increase in the rate at which the wire cuts the field, but this isn't to due with Faraday's law which relates the change in flux linkage on a 2-dimensional conductor. **the students** understandable reasoning for his answer of an induced EMF of 0 came from him observing that the flux linkage is of constant magnitude, so will have 0 rate of change, no matter the acceleration. Using the left hand rule, and the fact that magnetic force, F, is the sum of the BQv, it can be seen that the electrons will be moved by a greater force, making for a greater potential difference, EMF or current, as the wire accelerates.

    I also realised that with constant acceleration the voltage would be constant, if this was to do with Faraday's law, and will add that to what I have say.

    I'm asking as I want to help this person, while not feeding them false information, wrongfully undermining my teacher and ultimately embarrassing myself.
     
  2. jcsd
  3. Feb 13, 2017 #2

    cnh1995

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    Motional emf Blv is nothing but Faraday's law (the flux rule) in disguise. Emf induced in a moving wire in a uniform field is nothing but the product of magnetic field*area swept by it in one second=B*dA/dt.
    In case of a solenoid, the solenoid is not moving, but the flux is changing with time.
    So, E=A*dB/dt
    Both these equations ultimately give E=dΦ/dt, where B is constant in the first case and A is constant in the second case.

    You'll find a rigorous mathematical proof in the wikipedia article about this.

    Flux rule E=dΦ/dt is always true, be it a stationary circuit or a moving circuit. In case of a moving circuit, using Blv instead of dΦ/dt is simpler. You can see that Blv=Bl*dx/dt. Now ldx is the infinitesimal area dA swept by the conductor in time dt. So we can write it as BdA/dt which gives dΦ/dt(Φ=BA).
     
  4. Feb 14, 2017 #3
    Apologies, I've never come across the equation emf = blv. I instead referred to the equation F = bqv, for an electron, and how the wire is made up of these electrons.

    The differentiations made little sense to me, do you think you could explain it starting with the knowledge I've spoken about, and then perhaps introducing things like emf = blv etc.?

    Also, who was correct, the teacher?
     
  5. Feb 14, 2017 #4
    I think i understand now, because the wire is accelerating, the rate of change of the area swept by the wire is increasing, so faradays law works, as flux linkage is proportional to B and A, not just B.

    Is the explanation using lorenzt force invalid then?
     
  6. Feb 14, 2017 #5
    Again, with the constant acceleration of the airplane, wouldn't the change in flux linkage be constant, making for a constant voltage as opposed to an increasing one?
     
  7. Feb 14, 2017 #6

    cnh1995

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    Right.
    E=Blv follows from F=qVB on the electrons.
    Induced current will be zero if there is no closed path. If there is a closed path, you still need to specify the resistance of the loop. So I think the question should be 'what will be the induced emf?'.
    Force on the electrons F=qVB. This pushes the electrons on one side of the conductor. Due to the separation between electrons and protons, an attractive electric force F=qE acts on the electrons in the opposite direction to that of the magnetic force, where E is the electric field caused by the separation of charges.. This process stops when Lorentz force is equal to electric force i.e. qvB=qE.
    This gives E=vB. Potential difference (or emf) across the ends of the conductor is simply electric field*distance between the ends.
    Hence, induced emf Eind=vBl.
    (length of the conductor is the distance between its ends).
     
  8. Feb 14, 2017 #7

    cnh1995

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    No. Constant acceleration means increasing velocity. The emf will be increasing.
     
  9. Feb 14, 2017 #8
    in this situation = -d(BA)/dt = -B dA/dt = -BL dv/dt, right?

    Won't the different in velocity be constant, due to constant acceleration
     
  10. Feb 14, 2017 #9

    cnh1995

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    No. dA=ldx.
    v=dx/dt.
    So dA/dt=ldx/dt= vl.
    For constant acceleration, you'll need to consider instantaneous velocity.
    The equation will then be,
    Eind=Bl*v(t)
    Where v(t)= v0+at
    v0 is the initial velocity of the conductor and a is the constant acceleration.

    This equation is dimensionally incorrect too. This is for dEind/dt and not Eind.
     
  11. Feb 14, 2017 #10

    cnh1995

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    It appears to me that your teacher was correct.
     
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