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Falling square loop

  1. Dec 24, 2011 #1
    1. The problem statement, all variables and given/known data
    this comes from griffiths 7.46/11

    A square wire loop is placed so that its top portion is in a uniform magnetic field. It is then allowed to fall under gravity. We then repeat this experiment but instead rotate the loop 45° so that it is a diamond.

    Which orientation falls faster? Find the ratio of the terminal velocities.

    Finally, if the loop is allowed to freely rotate, which orientations would it assume while falling?

    2. Relevant equations

    Magnetic Flux: [itex]\Phi = \int{\vec{B} \cdot d\vec{a}}[/itex]
    Induced emf: [itex]\epsilon = -\frac{\partial \Phi}{\partial t}[/itex]
    Induced current: [itex]I=\epsilon/R[/itex] where R= resistance
    magnetic force: [itex]F_{mag}=I\int{d\vec{l} \times\vec{B}}[/itex]

    3. The attempt at a solution

    1) Find the flux and therefore the induced emf/current

    2) Calculate the magnetic force and set it equal to mg to find the terminal velocity [itex]F=mg=F_{mag}[/itex]

    I neglected the actual fomulas/ratio because it's mostly just math which isn't my difficulty.

    We basically end up finding that the diamond loop falls faster until the "half way mark" at which point the square loop falls 2x faster. After this, the diamond loop falls faster again.

    My difficulty is trying to figure out which orientation the loop takes on while its falling. The solutions manual says that it initially takes on the diamond orientation before switching to the square orientation at the half way mark. The manual says that this is because it attempts to "present the minimum chord at the field's edge" but I don't quite understand this explanation.

    I tried to rationalize it with momentum conservation in that the loop attempts to conserve velocity since its mass is constant, so it orients itself to the square shape but I don't really know if that applies here.

    Any help is appreciated! thanks.
    Last edited: Dec 24, 2011
  2. jcsd
  3. Dec 25, 2011 #2

    rude man

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    The problem is not clearly posed. For example, whether the B field is parallel to the loop's normal. We'll assume it is.

    Then, we have to further postulate. The only way this problem and its solution as stated make sense is if the bottom, not the top, of the loop touches the B field, which extends everywhere below the loop. Then the loop area will see a B field increasing with speed of fall until the whole loop is immersed in the B field.

    The B field could also be assumed to be finite below the top of the loop and extend down to somwhere like the bottom of the loop, but not all the way to the floor.

    The idea is we need to see dφ/dt somewhere along the line here.

    We'll assume a uniform B field from the loop bottom (initial position) to the floor.

    Under these circumstances a current will be induced as the loop falls, and a force acting on the loop will ensue in consequence, maybe also a torque, exactly per you equations.

    Emf generated around the loop will be = -dφ/dt where φ = φ(t) as the loop falls into the B field. This quantity will obviously be affected by the initial orientation of the loop. Then current I = emf/R.

    Then use the differential form of the Biot-Savart law to determine the force on each element of the loop dl as it enters the B field. This will be trivial for the square orientation but not for the diamond shape:

    dF = I(dl x B) along each side.

    In the case of the loop initially being a square I think you'll find there is no net torque applied about an axis at the loop's center and parallel to the B field. In the case of an initial diamond shape it might be different. If there is a net torque in that case then you'll get rotation of the loop about its axis until the torque is zero.

    The rate of fall is totally a function of the force on the loop, i.e. gravity minus magnetic force.
  4. Dec 25, 2011 #3
    Apologies, for that, I did not copy the question verbatim. B points into the page and occupies the top portion of the loop. However, this is similar to your situation where it points out of the page but takes up the bottom portion.

    The question is a bit vague when it says it's allowed to freely rotate but I'm guessing it assumes an initial diamond orientation. How would we calculate the torque for this?

    And is it necessary? The solutions manual doesn't use any calculations to achieve its answer so I feel like there is a more conceptual/intuitive explanation.
  5. Dec 26, 2011 #4

    rude man

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    There are no calculations because there are no numbers associated with the problem. But you still have some input: square loop, alternatively inserted into the B field in horizontal/vertical and in diagonal ("diamond") orientations.

    You need to compute dφ/dt for both cases as the loop accelerates into the B field due to gravity. That gives you the current I by your stated formulas. Then the force on each side is F = I(l x B). In the diamond case you have to resolve these forces, which act at 45 deg. to the vertical on all four sides, to determine how much more or less net vertical force there is compared to the rectangular case. (For exdample, there is obviously no vertical force on the vertical sides of the loop, but there is if inserted diagonally).

    Then you have to see whether there is a net torque applied to the loop as it enters the B field, about an axis thru the loop's center and parallel to the B field.

    If you want further help from me I ask that you copy and post the problem verbatim. Thanks.
  6. Dec 27, 2011 #5
    Shouldn't the nonvertical forces cancel out in both the diagonal and square cases though?

    And here is the problem, verbatim.

    7.46) Refer to Prob 7.11
    (a) Does the square ring fall faster in the orientation shown (Fig 7.19 [attached]), or when rotated 45° about an axis coming out of the page? Find the ratio of the two terminal velocities. If you dropped the loop, which orientation would it assume in falling? [Answer: (√2-2y/l)^2 where l is the length of a side, and y is the height of the center above the edge of the magnetic field, in the rotated configuration.]

    (b) How long does it take a circular ring to cross the bottom of the magnetic field, at its (changing) terminal velocity?

    A square loop is cut out of a thick sheet of aluminum. It is then placed so that the top portion is in a uniform magnetic field B, and allowed to fall under gravity (Fig 7.19). (In the diagram, shading indicates the field region; B points into the page.) If the magnetic field is 1 T (a pretty standard laboratory field), find the terminal velocity of the loop (in m/s). Find the velocity of the loop as a function of time. How long does it take (in seconds) to reach, say, 90% of the terminal velocity? What would happen if you cut a tiny slit in the ring, breaking the circuit? [Note: The dimensions of the loop cancel out; determine the actual numbers, in the units indicated.]

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  7. Dec 27, 2011 #6

    rude man

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  8. Dec 29, 2011 #7
    Ah. Thank you. That definitely helps a lot.

    The solution from the manual is:
    "If free to rotate, it would start out in the 'diamond' orientation, switch to 'square' for the middle position, and then switch back to diamond, always trying to present the minimum chord at the field's edge"

    I'm not sure, but perhaps the problem means if the loop is free to rotate to the lowest energy?
  9. Dec 29, 2011 #8

    rude man

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    From the wording of the answer, it would seem that the loop did not start in the middle position, where the loop is half-covered by the B field, as shown in the diagram. In other words, the problem was not clearly stated, as I see it. If the loop starts to fall higher than shown in the diagram, then the bottom sides (for the diagonal position) would also experience a force.

    I would not worry too much about getting the answer they sought, in other words. Just be clear about how an emf is generated around a loop (dφ/dt), and the equivalent statement of how an emf is generated in a section of wire dl which is (v x B)*dl and how a B field acts on the resulting current I = emf/R and dF = I(dl x B). I.e. just the equations you properly cited at the outset, except keep in mind that in the diagonal case the force F is exerted only on that part of each side of the loop immersed in the B field.
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