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The Magnetic Field

  1. The Biot-Savart law allows us to compute the (vector) magnetic field as
    B = (μ / (4 π)) I ∫ dl ⊗ r / r3,
    where (again, for a uniform medium) μ is the magnetic permeability and μ0 (for a vacuum) ≡ 4 π * 10-7 T m / A. Note that the form of the Biot-Savart law is the same as Coulomb's law with the substitutions

    As with the electric field, r is a vector pointing from the current element I dl to the field point.

    Note that the direction of integration must be in the direction of current flow.
    The direction of B can be given by another right hand rule: place the thumb of your right hand along the current; the field will be tangent to the curved fingers of your right hand. In this figure, the black arrow is the current and the red and blue arrows show the direction and magnitude of the magnetic field at the base of each arrow:

    Combining this with

    F = It ∫ dlt ⊗ B
    we see that parallel currents attract, and anti-parallel currents repel.

    Consider a current element with current I and length b, flowing from the origin at an angle θ in the x-y plane. The field point is (x,y,0), the source is parametrized by s and dl = (cos θ, sin θ, 0) ds:

    r = (x - s cos θ, y - s sin θ, 0)

    dl ⊗ r = (cos θ ds i + sin θ ds j) ⊗ ((x - s cos θ) i + (y - s sin θ) j)

    = ((y - s sin θ) cos θ - (x - s cos θ) sin θ) k ds

    = (y cos θ - x sin θ) k ds

    r = √ (s2 + x2 + y2 - 2 s (x cos θ + y sin θ))
    Using the same substitution we used for the electrical potential:
    u = s - (x cos θ + y sin θ)
    = s - a
    we have
    B(x,y,0) = (μ I (y cos θ - x sin θ) k / (4 π)) ∫- ab - a du / (u2 + x2 + y2 - a2)3 / 2
    = (μ I (y cos θ - x sin θ) k / (4 π)) u / ((x2 + y2 - a2) √ (u2 + x2 + y2 - a2)) |- ab - a

    = (μ I k / (4 π)) ((x cos θ + y sin θ) / √ (x2 + y2) +

    (b - (x cos θ + y sin θ)) / √ (b2 + x2 + y2 - 2 b (x cos θ + y sin θ))) /
    (y cos θ - x sin θ)
    In a similar fashion, we can compute the field due to a current element from (-b,0,0) to (b,0,0) as (note the negative sign required for the correct current flow)
    B(x,y,0) | θ=0 - B(x,y,0) | θ=π
    = (μ I k / (4 π y)) ((b - x) / √ ((b - x)2 + y2) + (b + x) / √ ((b + x)2 + y2))
    which in the limit b → ∞ is
    B = μ I k / (2 π y)

  2. A Helmholtz Coil consists of two loops of radius r (N turns each), each coaxial with the z axis, and centered on z = -r/2 and z = r/2. Taking advantage of the axial symmetry, we choose our field point to be (a,0,0) and our source point to be
    (r cos θ, r sin θ, b * r / 2)
    where b = +1 or -1 (we will do each loop separately).
    dl = (-r sin θ, r cos θ, 0)
    and
    dl ⊗ r = (-b/2 r2 cos θ, -b/2 r2 sin θ, r (r - a cos θ)).
    Integrating with respect to θ, from 0 to 2π, we obtain (for both loops together) a very complicated function, whose only nonzero component is in the z direction. But expanding in a Taylor Series about a = 0, we find that the result simplifies greatly, to
    B = 8 μ N I / (5 √ 5 r) k + terms of order a3.
    If we allow the field point to vary in z as well, the result is even messier, but numerically we can see that Bx and Bz are constant in a finite region between the coils (By is identically zero since we have computed B in the x-z plane):

    (8 / (5 √ 5) ≈ 0.7155; solid red (above) or green (below) represents 10% deviation from the central value.)

    So we see that the Helmholtz Coil can be used to generate an essentially constant magnetic field in a finite region.

  3. Ampere's law (another application of Stoke's theorem) is
    ∫closed loop B ⋅ ds = μ I,
    where I is the total current passing through the closed loop (imagine the loop as the boundary of a disc; the currents that contribute pierce the disc, with positive sign if the current is parallel to the normal of the disc as determined by the right hand rule).

    These "Amperian loops" are often useful:

    For both the toroidal coil and the solenoid, B is strictly constant inside the coil and zero outside only in the ideal limit where the individual turns of wire can be considered tightly packed circular current loops (and in the case of the solenoid, an infinitely long coil).
    As with Gauss' law, the Amperian loop must intersect the field point. Also, the loop must be far from ends or corners of a real current.
    Use the right hand rule to decide the direction of B, and where it is zero and nonzero.
    We should also ask the question "how far from the ends is far enough?"

    It turns out the calculation is virtually identical to the one we did for a linear electric charge, with the same result: for a straight current element whose ends are 10 times further away than the field point near the center, the error is again 0.5 %.

  4. We define the magnetic flux:
    ΦB ≡ ∫ B ⋅ dA.
    Because there are no point-like objects with magnetic "charge", Gauss' law gives us
    ∫closed surface dΦB = 0.

  5. The magnetic field can also be described in terms of a potential, but it is a vector potential, and to do so we need two new operators, the divergence:
    ∇ ⋅ A ≡ dAx/dx + dAy/dy + dAz/dz,
    and the curl:
    ∇ ⊗ A ≡ (dAy/dz - dAz/dy, dAz/dx - dAx/dz, dAx/dy - dAy/dx).
    With these definitions,
    B = ∇ ⊗ A.
    Gauss' law can be written as
    ∇ ⋅ B = 0,
    and is then a result of the identity
    ∇ ⋅ (∇ ⊗ A) ≡ 0.

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©2010, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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