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Homework Help: Magnetic energy problem

  1. Jan 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A small circular loop, radius a in the xy plane, carries current I which creates a magnetic moment pointing in the z direction. A magnetic induction field B= (s/y^3)z is applied to the loop where s is a constant, y is the distance in the plane of the loop from an arbitrary origin to the centre of the loop and z is a unit vector in the z direction. Assuming that a<<y, show magnetic energy is.

    [tex] U = -\frac{s}{y^{3}} ( \pi a^{2} I [/tex]

    hence determine the force on the loop in magnitude and direction.

    2. Relevant equations &
    3. The attempt at a solution

    from what i know,

    magnetic dipole m = I A
    [tex] m = I \pi a^{2} [/tex]
    induced magnetic field is in the positive z direction, i dont understand how the distance y comes into this though, magnetic fields fall off quickly so i cant see why itd be used..
  2. jcsd
  3. Jan 8, 2009 #2


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    They simply give you B, and you have calculated m. Now, you need the simple expression for the configuration energy of a magnetic diople in a magnetic field.
  4. Jan 9, 2009 #3
    Ok i had a minore brain fart, i wasnt thinking of the magnetic moment to arrisen from a dipole for some reason (even though i know thats how they arrise). So energy is simpley given from U = - m.B which would give me the equation in the question as angle is 90 degrees.
    Now force:
    [tex] F = - \nabla U [/tex]

    bu im assuming that the arbitrary distance y here can be treated as a constant?? therefore no diferentiation, in which case i have a2 in polar co-ords as p2 do i use [tex]cos(\theta)[/tex] in the expression for U (energy) to get its functional dependance on U? im guessing i do. then i have to get nabla in polar co-ords, where:

    [tex] F = -( \frac{\partial U}{\partial \rho} \hat{e}_{\rho} + \frac{1}{\rho} \frac{\partial U}{\partial \theta} \hat{\theta} [/tex]

    am i right?

    edit: im pretty sure im incorrect as im using polar co-ords but the angle is inbetween 2 vectors and they are pointing through the z plane (x-y plane). :S do i use cylinderical/spherical co-ords?
    Last edited: Jan 9, 2009
  5. Jan 11, 2009 #4
    Can no-one be of asisstance? I think also the problem is with the origin and loop centre being displaced by a large distance (in relation to radius) this makes the method slightly less trivial.
  6. Jan 11, 2009 #5


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    Hmm... Since the potential depends on the position of the loop. one would expect, in general, that if the loop were moved to a different location in the magnetic field, the force on the loop would be different. So.... let's take a look at which quantities in your energy expression actually vary as the loop is moved around in the magnetic field:

    [tex]U = -\frac{s}{y^{3}} ( \pi a^{2} I )[/tex]

    You are told that [itex]s[/itex] is just a constant; so that doesn't change. There is no reason to expect radius of the loop to change as the loop is moved around in the field, so you can treat [itex]a[/itex] as a constant. The current in the loop is assumed to be independent of the loop's location in the field, so [itex] I[/itex] is also constant. That leaves you with [itex]y[/itex] the distance to the center of the loop from the origin. Clearly that changes as the loop is moved correct?

    Now, how to make use of that?....For starters, I would go ahead and use polar coordinates [itex]\rho[/itex] and [itex]\phi[/itex]. Since [itex]y[/itex] varies as the loop is moved through the field, you would expect it to be a position dependent quantity and so in polar coordinates it is some function of [itex]\rho[/itex] and [itex]\phi[/itex] (i.e. [itex]y=y(\rho,\phi)[/itex])

    Now in order to find the force on the entire loop, you could first find the force on each infinitesimally small piece of the loop and then add them all up. Or you could recognize that since [itex]a\ll y[/itex], the force isn't going to vary much over the loop; that means you can effectively consider the loop as being located entirely at its center when calculating the force....So in that case [itex]y(\rho,\phi)=???[/itex]:wink:
  7. Jan 11, 2009 #6
    Im not sure about the polar co-ords,
    [tex] U = - \vec{m}.\vec{B} = - mBcos\theta [/tex]
    but the angle between m and b is at right angles to the xy plane,
    it is asked to calculate the potential of the plane at some point y, but i dont see how i could use polar co-ords as they are for 2 dimensions and [tex] theta [/tex] is not in the plane of radial distance, [tex]\rho[/tex]
  8. Jan 11, 2009 #7


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    The current loop is in the xy-plane, so the dipole moment it creates points in the z direction (right hand rule). It is parallel to B.

    Technically it is still a 3D problem, so if it makes you feel better you can use cylindrical coordinates [itex]\rho[/itex], [itex]\phi[/itex] and [itex]z[/itex]; but [itex]U[/itex] has no z-dependence so it really makes no difference.
  9. Jan 11, 2009 #8
    So i dont need the functional dependance of U on Theta (for the purpose of differentiating)? or i do? because that is what is confusing me, for polar co-ords the angle theta is in the xy plane but here its perpendicular to xy.
    For the future, how would i know when or when not to use polar co-ords in 3d problems?
  10. Jan 11, 2009 #9


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    Of course you do; but you should be able to reason that in this case, [itex]U[/itex] has no dependence on theta (the polar angle)

    Huh? why do you think that theta is perpendicular to the xy plane in this case?:confused:

    Draw a picture! Label the coordinates of the center of the loop using cylindrical coordinates; you should see clearly that the 'y' they refer to in the question is the same thing as the cylindrical coordinate [itex]\rho[/itex] Remember, in this case 'y' is defined as the distance from the origin to the center of the loop


    [tex]U = -\frac{s}{\rho^{3}} ( \pi a^{2} I )[/tex]

    and you can find the force on the loop by taking the gradient as usual.

    Remember, this entire problem rests on you treating the small loop as a pure dipole located at the center of the loop.
  11. Jan 11, 2009 #10
    Sorry badly really worded on my part, i meant to say that B is perpendicular to the plane/area created by the magnetic dipole.
    If the distance from the centre of the loop (on the plane xy) to the point y is rotated around centre of loop by 360 degrees it creates a circle on plane xy with ring being distance y away from it, but this plane is perpendicular to the surface normal (of the loop of wire area; pi R^2) which is paralell to the field B, i see the problem in using polar co-ords as the angle is measured from the surface normal to some field direction B, (in this case its paralell to surface normal) so basically this angle is not measured in the xy plane and if we were to use polar co-ords wed need both the angle flat on the xy plane, with the point p (at distance y) on the xy plane (which it is)
    sorry if it was confusing, found it a little hard to explain what i meant:S
  12. Jan 11, 2009 #11


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    Careful....B is perpendicular to the surface enclosed by the loop, but so is the normal to that surface; and the direction of m is in the direction of the normal. Hence, m is parallel to B.

    Huh? this is just as confusing.

    [itex]\rho[/itex],[itex]\phi[/itex] (or theta) and [itex]z[/itex] are measured from the origin, not from the center of the loop.

    Since [itex]U[/itex] depends only on the distance of the loop from the origin, it is [itex]\phi[/itex] and [itex]z[/itex] independent.
  13. Jan 12, 2009 #12
    ok done :)
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