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Homework Help: Inductance problem: mass dropped above magnetic field

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data
    "Consider a square metal picture frame of side length s, mass M, and total electrical resistance R. It is dropped from rest from a height H above a region of uniform magnetic field pointing into the page. The frame accelerates downward under the influence of gravity until reaching the magnetic field. It is observed that, while entering the magnetic field, the frame moves with constant velocity. What is the frame's speed when it begins to enter the magnetic field? What is the strength of the magnetic field? (Solve for B in terms of the given physical quantities (s, M, R, and H) and any constants you need.) After the frame has completely entered the magnetic field, what is the frame's acceleration? Justify your answer in two sentences or less."


    2. Relevant equations
    [tex] {V_{f}}^2 = {V_{i}}^2 + 2 a x [/tex]
    [tex] F_{mag} = \frac{v l^2 B^2}{R} [/tex]
    [tex]F = m a [/tex]

    3. The attempt at a solution
    First I found the velocity after the metal frame dropped a height H:

    [tex]V_{f}^2 = V_{i}^2 + 2 a x [/tex]
    [tex]V_{f} = \sqrt{ 2 a x } = \sqrt{ 2 (9.8 \frac{m}{s^2 }) H} = 4.4 H \frac{m}{s}[/tex]

    Then I reasoned that the only portion of the square metal picture frame that would factor into the magnetic force was the bottom horizontal portion. I figured that the two sides were moving in the same direction as velocity, and therefore the cross-product would go to zero. I also reasoned that the top horizontal portion would not matter, because as soon as this portion reaches the magnetic field, the flux is no longer changing and there will be no induced magnetic field, and therefore no magnetic force.

    I approached the next portion as a dynamic equilibrium problem. Since the velocity is constant, I summed the forces in the positive and negative y-direction. Downward I have [itex]F_{gravity} = m g[/itex], and upward [itex]F_{B} = \frac{v s^2 B^2}{R}[/itex]. I then set them equal and solved for B, obtaining:

    [tex]B = \sqrt{ \frac{M g R}{4.4 H s} }[/tex]

    After it completely enters the field, I assumed that its acceleration would return to [itex]9.8 \frac{m}{s^2}[/itex] because the flux is no longer changing, so no fields are being induced.

    Is this at all on the right track?
  2. jcsd
  3. May 9, 2015 #2

    rude man

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    Vf and B are both wrong. Redo, leaving all parameters in terms of symbols rather than numbers until the very end. You are missing chances to check dimensions term-by-term as you go. You got all the right relations and then blew it by introducing numbers prematurely.
  4. May 9, 2015 #3
    dude, that's almost what i got, except i left gravity as g, and for s, i considered the length of the whole perimeter to be 4s, then squaring I got 16s. are you in simpson's class? anyways, i used the conservation of energy equation for part A though: MgH = 1/2 Mv^2
    Last edited: May 9, 2015
  5. May 9, 2015 #4


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    Hi monkeymind. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    You did a good job of explaining your working, which makes errors easy to spot.

    rude man makes a good point about not being too hasty in substituting data values for symbols in your formulae.
    Last edited by a moderator: May 7, 2017
  6. May 9, 2015 #5

    rude man

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  7. May 9, 2015 #6
    oops, sorry... 16s^2... is that wrong rude man ? I figured it would be right, because this is the whole length in which current can flow through.
  8. May 9, 2015 #7

    rude man

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    That's the area, not the length. The length is 4s. But I'm not sure why you are interested in it.
    Monkeymind did a good job with his reasoning, you should look at it. Just remember not to switch to numbers until the end.
  9. May 9, 2015 #8
    @rude man , I did the same thing monkeymind did, but i didn't plug in numbers. and I didn't use the area, I used the length of the whole picture frame (if it was straightened out, it would be 4s). the formula for magnetic force is F = V(SB)^2/R = V S^2 B^2 / R so instead of plugging in just s, I plugged in 4s.
  10. May 9, 2015 #9

    rude man

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    Problem is, @PhysicsChick, that only one of the four sides gets the mag. force in the vertical direction.
  11. May 9, 2015 #10
    OK, thanks for the help, I really appreciate it! That actually makes sense now that I think about it lol.
  12. May 9, 2015 #11
    I see that I made two mistakes (hopefully that's it!). One when calculating vf and one when calculating B.

    Correcting, this gives me:

    [tex] v_{f} = 4.4 \sqrt{H} \frac{m}{s}[/tex]
    [tex] B = \sqrt{\frac{MgR}{(4.4\sqrt{H})s^2}} [/tex]

    Is that better?
  13. May 9, 2015 #12
    I got [tex] B = \sqrt{\frac{MgR}{(\sqrt{2gH})s^2}} [/tex]
  14. May 9, 2015 #13

    rude man

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    That is right, and is probably what monkeyminbd would have gotten had he/she stuck to symbols to the end as you did.
  15. May 9, 2015 #14
    I'm sure it's the same thing.
  16. May 9, 2015 #15
    Thanks for your help, rude man! I will definitely start solving symbolically from now on.
  17. May 10, 2015 #16

    rude man

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    I don't want to over-dramatize, but that is one of the most important things to learn in your physics studies.
    So now try also to check units in each term. I prefer the term "dimensions". Like, force would have dimensions {F} = MLT-2, M for mass, L for length, T for time. When you get into elec. & magnetism you have to add Q for charge. You can deduce the dimensions of B for example by the familiar formula
    F = iLB so B = F/iL and so {B} = {F}/{i}(L} = MLT-2/QT-1 = MLT-2Q-1TL-1 = MT-1Q-1.
    Or, from ampere's law B = μi/L, {μi/L} = {B} which means that {μ} = {B}{L}/{i} = MLQ-2.
    Last edited: May 10, 2015
  18. May 10, 2015 #17
    @rude man that's what we're taking right now! lol. Thanks for your help! Much appreciated =]
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