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Effects of Magnetic field span reduction?

  1. Aug 5, 2015 #1
    Given the following diagram:
    The conductor(C) of known volume(##V##) and resistance(##R##), passes a uniform magnetic field(##B##)at a constant velocity(##v##), via the following formula: ## \epsilon = -vBL## the induced EMF can be calculated. This conductor acts as a power source to a certain circuit,therefore, it's connected to a circuit(which I chose not to include in the diagrams because im focusing on this conductor specifically) to distinguish the resistances I'd like to label ##R## as the internal resistance of the power source( that is the conductor) and r as the resistance of the circuit.

    I think the way of calculating the induced current here is simply Ohm's law:

    ##I = \frac{V}{R + r}##

    the total resistance is it ##(r+R)##, not sure though...

    Now, the main point, what would be the effect of reducing the magnetic field span(Magnetic field span: the area the magnetic field covers at the same strength) like so:

    While ##L##, ##v## , and ##B## are all the same? My assumption: that the induced EMF is the same, however, the second diagram/version would induced a lower current? But why...?

    It makes no sense to me that the the currents would be the same magnitude, it does in-terms of EMF because of all the variables in the motional EMF formula are the same.
  2. jcsd
  3. Aug 6, 2015 #2


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    Hey there!
    No reply yet huh? Well, I'm going to chime in then! Please take my comments with a grain of salt as I am here to discuss it with you rather than give you an answer (simply because I don't think I have one! I'm just bored at work o0)).

    There are two things I am thinking about:
    With the first figure, there will be an applied voltage difference (or emf I should say) along the long sides of the moving conductor (all the way). For the second figure, there will only be a voltage difference (emf) at the part of the moving conductor that is with in the small magnetic field (only part of the long sides). For the second figure, this applied voltage (emf) will travel along the long sides of the moving conductor whereas for the first image it is set throughout the long sides. What do you think?

    You say that this moving conductor is connected to a circuit. I think the circuit that you speak of is needed to determine the result. Specifically speaking I think the point where the contacts meet the moving conductor is important. What do you think?

    If you get an answer please post!
  4. Aug 6, 2015 #3
    Hi there!
    Well, you could simplify things, we could assume that the induced EMF is only along the side with respect to the magnetic field, which makes sense. However, since all the variables are the same(via motional EMF law) the induced EMF in both diagrams should be the same(with respect to ## v,B,L##). I think the induced current would not be equal, based on the point of the magnetic field span being smaller and the point you made too.

    As for the circuit here is a simple diagram of it:


    Where (C) is the conductor we're talking about.

    @jim hardy as always your opinion please.
    Last edited: Aug 6, 2015
  5. Aug 6, 2015 #4


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    Yes, I agree with you, current would increase when arranged as figure 1. I think this is because figure one can be thought of as multiple figure 2s in parallel (multiple emf sources connected in parallel). So i am think of it as comparing 1 battery (fig2) with n batteries connected in parallel (fig1). Fig1 and Fig2 will have the same emf, but since there are multiple "emf sources" in fig1 (as per my previous crude statement), they will provide more current (assuming the contact is all along the long sides of the conductor). I think that R (resistance of the conductor) changes according to the configuration (fig1 vs fig2). Maybe if you have resistivity (and you have the volume) you can calculate the resistance of the conductor?

    Yes, hopefully Jim Hardy will swoop in for the save (or any PF vet for that matter).
  6. Aug 6, 2015 #5
    That does make sense when we compare it to ta battery in parallel.
    In regrades to the resistance of the conductor I have calculated some random values you'd get the same resistance all around.
    Since the volume of the conductor stays the same, and resistivity of the material is a given constant... resistance in Fig1/Fig2 should be the same.

    That's why I tag Jim, always gives the goods!
  7. Aug 6, 2015 #6


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    I think the resistance does change (key word here is "think" lol), if you use only the area under the magnetic field, this area would change according to fig1 and fig2. Since the effective part of the conductor that is conducting changes according to each figure. This is of course assuming you have the dimensions of the conductor and the magnetic field in fig2 along with the resistivity. Once again, I am not sure of all this, just curious and enjoy discussion. Yes, Jim is one of a kind ; ) PF in general is a wonderful place.

    EDIT: I am not sure of this actually, I am probably wrong about this point. Apologies for this. I will keep this post regardless.
    Last edited: Aug 6, 2015
  8. Aug 6, 2015 #7

    jim hardy

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    in what plane do you envision that length term L ?

    Here's what i see for your first picture


    e = BLV , ignoring external circuit

    sorry, i cant do the fancy script

    and for your second picture


    point of connection for voltmeter leads is obviously significant.
    Voltage across other end will be smaller.
  9. Aug 6, 2015 #8
    So my initial

    They way you see it it's correct.
    Wait, is e = vBL is incorrect? I'm not sure as to how the voltage is divided up due to the total resistance of the conductor(area in/out of B).
    I assumed from e = vBL they'd be the same considering all variables being equal(vBL). Therefore, there is no way for it to be indeed equal, hence reducing the magnetic field span does indeed reduce the induced voltage therefore induced current.
  10. Aug 6, 2015 #9

    jim hardy

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    does it reduce induced voltage, or does it unblock parallel current paths as shown by the red lines ?
    Thereby allowing Kirchoff to do his work ?

  11. Aug 6, 2015 #10
    Voltmeter would be connected same area you initially labeled.
    Would the formula now be:

    ##V_E * \frac{R}{R_O +R}##

    Where ##V_E## is the induced EMF via (vBL) and R is the resistance of the portion inside the magnetic field ##R_O## portion of resistance outside the magnetic field?

    While the first conductor's induced EMF is simply ##V_E##?
  12. Aug 6, 2015 #11

    jim hardy

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    i think numerator would be ROutside the magnetic field

    voltage divider action... KVL....
  13. Aug 6, 2015 #12
    Yes, you're right on that one. I check back to the wiki of voltage dividers, ##Z_2## makes sense to be ##R_O##.
  14. Aug 7, 2015 #13
    @jim hardy , something is bugging me with the topic of voltage dividers. Let consider a change in the model I presented.
    Consider the same conductor. This time, exclude the magnetic field,motion and such. The induced EMF is from a power supply is connect to the conductor(C) a small portion of it like so:

    Would current #I## flow all around the conductive region? Or just the part that it's connected to? Would the concept of voltage divider be applied here?
    My assumption would say that the current would flow depending on the applied EMF polarity as diagram with the black arrow, but my gut feeling says otherwise, that it would flow in loops? Also, would the voltage divider still be valid? How would I apply it.
    Last edited: Aug 7, 2015
  15. Aug 7, 2015 #14

    jim hardy

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    You're a mechanical, right ?

    It's a field problem.

    How would stress distribute over the area between those points of connection?

    At any point in the orange rectangle there's a gradient between those points of connection.
    voltage at any point will be in proportion to its relative distance from the two points of connection
    and current will flow along the gradient lines, perpendicular to equipotential lines

    eg a straight horizontal line bisecting the rectangle will be an equipotential line at V/2
    the rest of the equipotential lines will be curves probably conic sections centered on the points of connection.
    We solved such problems graphically in my day because we took that course before we'd taken vector calculus.
    Draw your gradients....
    or set up some equations
    Last edited: Aug 7, 2015
  16. Aug 7, 2015 #15
    Yes, thankfully I switched from EE... I can't really depict things well relative to electromagnetic aspects.

    I remembered I posted this a while back, went there and reviewed it.
    Here is my assumption after that breif review and studying about equipotential lines relative to this problem.
    I adjusted the diagram here:

    I think the most important line is the black one(where the majority of current will flow in that direction). Even though current flow's all around the conductor as other members clarified me earlier, however, I could risk it and say that 90% of the current flow will be on that line. What do you think? Of course experimenting would be best to prove this, and from that last discussion I wanted to test it, but it's quite messy to do haha

    However, would current be equal ohms law? Should I consider the voltage being less due to the connection only being in a small part of the conductor?
    Is voltage division something I need to consider too?
    Last edited: Aug 7, 2015
  17. Aug 8, 2015 #16

    jim hardy

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    How did you arrive at 90 % ?

    Seems to me there'll be no current outside the conductor so i erased the right half of your red ellipse.

    Seems to me your conductor can be imagined as an some finite number of thin strands in parallel, i colored 'em yellow
    just like pipes in parallel except Ohm's law flow calculation does not involve square roots like Bernouli's


    you could add admittances of parallel strands if you calculated their lengths
    and figure out a series that'd evaluate to admittance of the orange block

    or you could assume the potential at any point along a strand, and hence any point on the orange plane, is proportional to its relative distance from the two corners where the connection is made

    and by finding the locus of all points where (distance to top right corner)/(distance to bottom right corner) = some constant ratio k
    you could write equations to define those loci
    thereby getting equations for equipotential lines
    and that's a plane geometry problem that gives something called Apollonius Circle which, for k = 1 has infinite radius hence the straight equipotential line at midplane.
    http://jwilson.coe.uga.edu/EMAT6680Fa09/KimB/EMAT6690/Essay2/6690 apollonius circle.pdf

    The other equipotential lines will be segments of circles
    and current flow will be perpendicular to them

    For someone fluent in vector calculus i'm sure there's a straightforward solution involving j, del, grad and so on.

    but back to the 90%
    length of your red line is not 10X that of your black line, so why would black line hog 90% of the current ?

    Tinker with the geometry.
    It's too much algebra for me, I'd sooner put ten amps through a piece of brass shim stock and map the voltage with a probe made from one of wife's sewing needles..
    Have to make a microvoltmeter, though. Hmmmm there's one for sale in my local junkshop...
  18. Aug 8, 2015 #17
    Hmm, that's just a hunch on based on the black equipotential line, I assumed the red ones would be the only ones there(excluding the one outside the conductor that I forgot to erase earlier...) I think the majority of current should flow there 90% was just a wild hunch...I'm still studying the rest of what you're saying, contemplating it in.
    If you do try that out, let me know of the results! Would be fun.

    I had something to share about this diagram:
    What IF the induced EMF's are the same, since(##v,B,L##) are , from the the voltage divider formula that mean's they are(due to ##Vin##). And instead found the current induced with respect to that small magnetic field?

    So assume, that ##V_i## is the same, but the induced current isn't because:

    ## V = IR_i ## therefore, ##I = \frac{V}{R_i}##

    ## R_i = \frac{\sigma}{l}##

    Where we calculated the relevant resistance in the small magnetic field ##R_in##?
    @vanhees71 like to know your opinion too please.
  19. Aug 9, 2015 #18

    jim hardy

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    What black equipotential line?
    Your orange conductor is encircled by a black line and if that black line is a Faraday cage there's no gradient anyplace inside it.

    the region in your field becomes a current source instead of a current sink
    so voltage between any two points in or on it = Vinduced - IR
    apply that along its edge and sketch current

    does this exercise have a goal ?
  20. Aug 9, 2015 #19
    The one I drew on post #15? There is a black line from the + to - that is what I meant.

    Yes, it does become a current source.

    What if, their induced voltages would be the same(always) however, when we consider induced current we take the relevant resistance with respect to the magnetic field?
    ##I = V * \frac{l}{\sigma}##

    Not directly no, studying all odd concepts that comes to my mind in electromagnetism, the same happened in terms of Physics but most of my "odd" questions/problems had answers here or other forms or in texts books, likewise with mechanical engineering relevant problems tend to figure things out. But applications/questions relevant to electromagnetic parts of physics I struggle greatly.
  21. Aug 10, 2015 #20
    Here is what I think @jim hardy , let's consider that we've added a static connection piece(equal to the conductor's width & thickness) top the top and bottom of the conductor that we we're talking about here:

    The polarity there(bottom +/ top -) is based off the induced EMF(##vBL##), while as, current flow is diagrammed there as (##I##), and that current flow would be equal to: ##\epsilon## ##*## ##\frac{l}{\sigma} ##
    I think the reduction of magnetic field does not have a direct effect on induced motional EMF(since v,B,L are equal) however, we look at the resistance inside of the magnetic field to calculate the current?
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