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Motional EMF and gravity

  1. Jun 5, 2017 #1
    1. The problem statement, all variables and given/known data
    7.11.png

    2. Relevant equations
    Flux rule: ξ= -dΦ/dt'
    F= ma

    3. The attempt at a solution
    Let's assume that the loop is going down with speed 'v <0'.
    Using Flux rule,
    ξ = - Bva
    Current will move in clockwise direction to increase Φ.
    The effect of magnetic force on AC gets canceled by the effect of magnetic force on BD.
    Net effect of magnetic force on the loop = Net effect of magnetic force on AB
    In CD, the velocity of charged particle is
    V = vx##\hat x## + v ##\hat y## : v<0
    = ##\frac{- Bva}{Rλ}\hat x ## + v ##\hat y## :
    where λ is charge per unit length
    R is resistance across the loop = resistance across AB

    Magnetic force on AB = λa(##\frac{- Bva}{Rλ}####\hat x ## + v##\hat y##)× B(- ##\hat
    z##)
    = λa(##\frac{- B^2 va}{Rλ}\hat y - vB(\hat x##))

    ##\frac{dv }{dt} \hat y = [\frac{-B^2a^2 v}{Rm} - g]\hat y##

    Solving the equation gives,
    v = ## \frac{-gRm}{B^2a^2} [1-e^{\frac{B^2a^2 t}{Rm}}##]

    vterminal = ##- \frac{gRm}{B^2a^2 }##

    For reaching 90% of vterminal,
    0.9 vterminal = vterminal##[1-e^{\frac{B^2a^2 t}{Rm}}]##
    solving it,
    t = ##\frac {-v_{terminal} ~ln(1.9)} g##

    t , here, depends on the dimensions while in question , it is said that it should be independent of dimension.
    So, what is wrong here?
     
  2. jcsd
  3. Jun 5, 2017 #2

    TSny

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    Hello.
    Does the symbol "a" represent the distance between points A and B in the diagram?

    It's not just the resistance of AB that determines the current in the loop, but the total resistance of the entire loop. When you say that λ is charge per unit length, I take it that means the charge per unit length of the free charge carriers.

    I think this is correct with R being the resistance of the entire loop as mentioned above.

    You didn't solve for t correctly.

    Can you express the mass m and the resistance R in terms of the mass density and resistivity of aluminum?
     
  4. Jun 5, 2017 #3

    scottdave

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    Maybe some will cancel? mass is related to the amount of aluminum and it's density {inherent property}. Resistance is related to the length and the cross section and the resistivity {inherent property}.
     
  5. Jun 5, 2017 #4
    Yes
    How do we know that it is resistance of the entire loop or resistance of AB?
    -0.1 vterminal=##-v_{terminal}~e^{\frac{B^2a^2 t}{Rm}}##
    ## ln(0.1)=\frac {-gt} v_{terminal} ##
    ##~t ~= \frac { -v_{terminal}~ln(0.1)} g ##
    ##\frac {-v_{terminal}}g= \frac{Rm}{B^2a^2 } = \frac{{\frac{ρ 4 a}{π r^2}}{μ 4 a}}{B^2a^2 } = \frac{16ρμ}{πB^2r^2 } ##

    Where μ is the mass density,
    ρ is the resistivity
    r is the radius of the wire

    So, I am getting r in the answer.
    What to do now?
     
  6. Jun 5, 2017 #5

    TSny

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    The induced emf drives current through the entire loop, not just between A and B.
    If ##\mu## is the volume mass density of aluminum, did you write ##m## correctly in terms of ##\mu##? Otherwise, your work looks good.
    [EDIT: The problem says that the loop was cut out of a sheet. So, the cross-sectional area of the loop would not be circular. You can just use ##A## for the area of the cross section instead of ##\pi r^2##.]
     
    Last edited: Jun 5, 2017
  7. Jun 6, 2017 #6
    I took μ as linear mass density which is wrong. I should take μ as volume mass density as the aluminum sheet has thickness,too.
    taking μ as volume mass density,
    ##\frac {-v_{terminal}}g= \frac{Rm}{B^2a^2 } = \frac{{\frac{ρ 4 a}{A}}{μ 4 a A}}{B^2a^2 } = \frac{16ρμ}{B^2 }##


    Done.
    Thank you.
     
  8. Jun 6, 2017 #7
    2nd part of the question,
    If I cut a tiny slit in the loop, there will be no current through the loop and so, the loop will free fall. Right?



    If I want to measure resistance of a loop using ohmmeter , I will put the terminal of the meter at two points of the loop , let's say A and B and hence I measure the resistance across A and B.
    What is the meaning of resistance across the whole loop? How can we measure it?
     
  9. Jun 6, 2017 #8

    cnh1995

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    Yes.
    You can't measure it directly when the loop is complete.
     
  10. Jun 6, 2017 #9
    So for measuring resistance of the whole loop, we have to break the loop and measure the resistance across the two end points.
    Is this the resistance of the whole loop?
     
  11. Jun 6, 2017 #10

    cnh1995

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    Yes.
     
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