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Homework Help: Magnetic Field at a distance from the center of a Torodial Solenoid

  1. Jul 1, 2008 #1
    1. The problem statement, all variables and given/known data


    A torodial solenoid has inner radius r1 = .15m, outer radius r2 = .18m; turns N=250 and carries a current I = 8.50A. What is the magnitude of the magnetic field at the following distances from the center of the torus?
    a) .12m
    b) .16m
    c) .20m

    2. Relevant equations

    B = [tex]\mu[/tex]0NI/2pir

    3. The attempt at a solution

    I'm actually not quite sure as to how to use the formula given the fact that we have inner and outer radii and then different distances on top of that.

    I was thinking the way to solve the problem would be to find the B fields at each of the radii and each of the distances given and then subtract given by the distance, so the B field at .12m would be subtracted from the B field at .15m, and than the other 2 would be subtracted from the .18m radius, like so

    B1 = 2.8*10^-3T
    B2 = 2.4*10^-3T
    B(.12) = 3.6*10^-3T
    B(.16) = 2.7*10^-3T
    B(.20) = 2.1*10^-3T

    B(.12)-B1 = 8*10^-4T
    B(.16)-B2 = 3*10^-4T
    B(.20)-B2 = 3*10^-4T

    not sure if I'm correct since my answers end in terms of a Gausse with such a small current, I was also thinking that the B field at .20m from the center would be 0 since it's outside the solenoid, if anyone could point out what I'm doing wrong I'd greatly appreciate it, thanks.

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  3. Jul 1, 2008 #2
    Can't you just use Ampere's law?

    Draw your amperian loop at the distance given by the question. For example for a (0.12 m), you draw a circle of radius 0.12m around the center of the coil and use Ampere's law.

    You can do the same for 0.16m and 0.18m.
  4. Jul 1, 2008 #3
    I did use Ampere's law.... B = (mu)NI/2piR, does this mean that the radii given in the problem introduction do not apply?
  5. Jul 1, 2008 #4
    You did not mention using Ampere's law once, you merely state the equation and not how you got it. You say you 'don't know how to use the equation', which means to me that you simply got it out of your book or something.
    The equation is derived from Ampere's law and during that derivation, the radius plays a huge role!
    The equation is only valid for a special value of R (do you know which?), but you will find that it is not valid for all radii!
    Think about it, think about how it is derived (from ampere's law) and try to do it yourself. Then try the same (seperately!) for the three radii you are given.
  6. Jul 1, 2008 #5
    well I know ampere's law states the the B field is only (mu)I, taken from the integral of B and dl, however around a circular path that integral also equals B(2piR), so it becomes

    B(2piR) = (mu)NI so B = (mu)NI/2piR, where 2piR is the circumfrence length of the circle which satisifies n = N/l = N/2piR, so it's the same as that of a regular solenoid B = (mu)nI, what I'm not sure is how the distances given interact with the inner and outer radii, I know Ampere's law has an application for cylindrical conductors (with radii of different lengths), what that be the correct equation to use

    B = (mu)I/2pir (with distances greater than radii)

    and B = (mu)Ir/2piR (with distances smaller than radii)

  7. Jul 1, 2008 #6
    You don't seem to understand Ampere's law very well.

    Ampere's law states that the path integral around a closed loop of the dotproduct of B and dl is equal to the enclosed current (by the loop) multiplied by mu_0:
    [tex]\oint \vec{B} \cdot \vec{dl} = \mu_0 I_{\text{enclosed}}[/tex]

    For a toroid, if you take a circular path through the center of the toroid (radius r), you will find that:
    [tex]B \oint dl = \mu_0 I N[/tex]
    Here, N is the number of coils in the toroid. Since the current passes through each of these coils the actual enclosed current is the current [itex]I[/itex] multiplied by the number of coils.

    The path integral over dl is just the circumference of the circle: 2 pi r. So:
    [tex]B 2 \pi r = \mu_0 I N[/tex]
    [tex]B = \frac{\mu_0 I N}{2 \pi r}[/tex]

    Now, do the exact same thing for the three questions (seperately, don't just find one formula and use it for all three!)

    For b), I think you can assume that the magnetic field is constant everywhere inside the toroid so it will be the same on 0.16m and on 0.165m (the center).

    For a) and c), think closely about the current that is enclosed! (Don't forget it's direction!)
  8. Jul 1, 2008 #7
    The correct equation to use is Ampere's law, not any of these you mention. They do not apply to every problem, because Ampere's law works out differently for every different situation. A toroid is certainly not the same as a cylindrical conductor, and therefore the 'equations' you get from Ampere's law are not the same either.

    (Sorry for double posting)
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