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B field from two coils

  1. Oct 14, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Helmholtz coils can be modelled as a pair of current loops oriented so that they are parallel to each other with a common axis in the x direction. The two loops have the same current I, and the same radius R, and their centres are at x = -d and x = +d, so the distance between the two loops is 2d.
    (a) Write down the magnetic field at position x along the axis of the coils. What is the magnetic field at the midpoint between the coils, i.e. at x = 0? What is the magnetic field at x >> R
    Sketch the magnetic field lines for all r.


    2. Relevant equations
    B field from a single coil

    3. The attempt at a solution
    Put the x axis horizontally with +ve x pointing rightwards. Then as you look in positive x, the loops have clockwise currents. The B field along the x axis for a single coil is $$\vec{B} = \frac{\mu_o I R^2}{2(R^2+x_1^2)^{3/2}}\underline{e}_x,$$ where ##x_1## is the B field at a distance x1 from coil 1.

    With the coils having the same orientation of current, both B fields will be in +ve x, so the total B field is given by: $$\frac{\mu_o I R^2}{2} \left(\frac{1}{(R^2+x_1^2)^{3/2}} + \frac{1}{(R^2 + x_2^2)^{3/2}}\right)$$ I can write ##x_1 = x - d## and ##x_2 = x + d##. Sub these in gives the first part. Sub x=0 for the second part: $$\vec{B}_{0} = \frac{\mu_o I R^2}{(R^2+d^2)^{3/2}}\underline{e}_x$$ For x >>R, taylor expand to give $$\vec{B}_{x>>R} = \frac{\mu_o I R^2}{2} \left(\frac{1}{(x-d)^3} + \frac{1}{(x+d)^3}\right)$$

    Is that ok? For the B field lines for all r, I have attached a sketch of what I have. I think the B field should be circular around the coil and obviously straight along the x axis. I am not too sure about the region between the two coils.
    Many thanks.
     

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  2. jcsd
  3. Oct 14, 2013 #2

    rude man

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    OK, altghough introducing x1 and x2 was of course unnecessary.

    Good work on the Taylor series expansion!
     
  4. Oct 14, 2013 #3

    CAF123

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    Hi rude man, thank you for the check. Do you agree with my sketch though? I know the divergence of B is always zero, so the B field should create loops.
     
  5. Oct 14, 2013 #4

    rude man

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    Yeah, I do. Nice work.
     
  6. Oct 14, 2013 #5

    TSny

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    Helmholtz coils are usually arranged so that the separation distance between the two coils is R. See http://en.wikipedia.org/wiki/Helmholtz_coil

    So, that would make d = R/2. Since you are assuming x >> R, that means x >> d. So, you can further simplify your approximation for this case.
     
  7. Oct 15, 2013 #6

    CAF123

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    Hi TSny,
    The distance between the two coils is usually taken to be R because then the second derivative of B vanishes => curvature of B field is zero (so direction is unchanged). This coupled with the fact that the first derivative vanishes => the B field is constant. Is this why the distance between them is R?

    The next part of the question asks to calculate the force between the loops. We are to use the fact that ##\vec{F} = \nabla (\underline{m} \cdot \underline{B})##

    ##\underline{m}## is the magnetic dipole moment of the loop so I can write ##\underline{m} = I \pi R^2 \underline{e}_x## Dot this with the B field compt gives $$\vec{F} = \nabla (\underline{m} \cdot \underline{B}) = \nabla \left(\frac{I^2 \pi R^4 \mu_o}{2(R^2 + (x+d)^2)^{3/2}}\right) = \frac{I^2 \pi R^4 \mu_o}{2} \nabla\left(\frac{1}{(R^2+(x+d)^2)^{3/2}}\right) = -\frac{3}{2}\frac{I^2\pi R^4\mu_o (x+d)}{(R^2 + (x+d)^2)^{5/2}}\underline{e}_x$$

    The force between the two loops is then this expression evaluated at x=d. Is this correct? I ask because we are a bit behind in lectures and we haven't actually covered why the force is given by the gradient of the divergence of these two quantities, so maybe I misinterpreted this. It makes sense that the force is in -ve x however.
    I can then sub d=R/2.
     
    Last edited: Oct 15, 2013
  8. Oct 15, 2013 #7

    TSny

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    Yes, the choice of d = R/2 makes the field very uniform near x = 0. The first nonzero derivative of the field with respect to x at x = 0 is the fourth derivative.

    Your calculation of ##\vec{F} = \nabla (\underline{m} \cdot \underline{B})## looks correct to me. However, I don't think this formula for the force on a dipole would be very accurate for the Helmholtz coils because the formula assumes that the currents making up the dipole moment are confined to a small region of space.
     
  9. Oct 17, 2013 #8

    CAF123

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    If I now consider the case where the current in one of the coils is reversed, then for z >>R (or equivalently z>>d) B will tend to zero. Now if I draw the field lines for this case, it will vanish for z=0 by symmetry but it is non zero everywhere else I think. How would I sketch this? See sketch. The B field from 'above' is downwards and that 'below' is upwards so they cancel at z=0. But how do I represent this on a diagram? - just a gap?
     

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  10. Oct 17, 2013 #9

    TSny

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    Yes. You won't be able to draw a field line along the z axis. Just forget that line. Same thing happens in other cases. For example, if you have two positive point charges where the electric field is zero at the midpoint between the charges.
     
  11. Oct 17, 2013 #10

    CAF123

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    The B field is not zero along the z axis though, right? (except at z=0). But because of this gap at z=0 it is convenient to just not draw the field line along z at all?
     
  12. Oct 17, 2013 #11

    TSny

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    Yes, the B field is zero only at z = 0. I don't see a consistent way to draw a field line running along the z-axis. The analogous case of the E field of two positive point charges is shown here. In figure (b) you can't draw a consistent field line along the line containing the charges.
     

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