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- The
**Biot-Savart law**allows us to compute the (vector)**magnetic field**as**B**= (μ / (4 π)) I ∫**dl ⊗ r**/ r^{3},**magnetic permeability**and μ_{0}(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- q → I
**dl**and - 1 / (4 π ε) → μ / (4 π).

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**= I_{t}∫**dl**_{t}**⊗ B**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)

r = √ (s**k**ds= (y cos θ - x sin θ)

**k**ds^{2}+ x^{2}+ y^{2}- 2 s (x cos θ + y sin θ))u = s - (x cos θ + y sin θ)

we have= s - a

**B**(x,y,0) = (μ I (y cos θ - x sin θ)**k**/ (4 π)) ∫_{- a}^{b - a}du / (u^{2}+ x^{2}+ y^{2}- a^{2})^{3 / 2}= (μ I (y cos θ - x sin θ)

**k**/ (4 π)) u / ((x^{2}+ y^{2}- a^{2}) √ (u^{2}+ x^{2}+ y^{2}- a^{2})) |_{- a}^{b - a}= (μ I

**k**/ (4 π)) ((x cos θ + y sin θ) / √ (x^{2}+ y^{2}) +(b - (x cos θ + y sin θ)) / √ (b

^{2}+ x^{2}+ y^{2}- 2 b (x cos θ + y sin θ))) /(y cos θ - x sin θ)

*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}+ y^{2}) + (b + x) / √ ((b + x)^{2}+ y^{2}))**B**= μ I**k**/ (2 π y) - q → I
- 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)**dl ⊗ r**= (-b/2 r^{2}cos θ, -b/2 r^{2}sin θ, r (r - a cos θ)).**B**= 8 μ N I / (5 √ 5 r)**k**+ terms of order a^{3}._{x}and B_{z}are constant in a finite region between the coils (B_{y}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.

**Ampere's law**(another application of Stoke's theorem) is∫

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)._{closed loop}**B ⋅ ds**= μ I,These "Amperian loops" are often useful:

- for axially symmetric currents,
**B**will be constant along and tangent to any given coaxial circle:A*long*straight current, field point a distance r from the current, in a plane perpendicular to the current:B * 2 π r = μ I

B = μ I / (2 π r)

- for a current wrapped around a torus,
**B**will be constant along any given circle coaxial with the torus:n loops per unit length, field point inside and coaxial to the toroidal coil, a distance r from the center of the figure:B * 2 π r = μ * n * 2 π r I

B = μ n I

If the Amperian loop is interior to the coil (between the center of the figure and the wires), the enclosed current is zero, and hence so is B. If the loop is exterior to the coil (beyond the wires), for every current element that crosses the plane of the loop in one direction, there is an identical current element crossing in the opposite direction, and so B is again 0.

- for a solenoid, use a rectangle parallel to the solenoid axis, with one side inside and one side outside: inside the solenoid,
**B**is constant and parallel to the solenoid axis, and outside the solenoid,**B**is zero:B *

*l*+ 0 * w + 0 **l*+ 0 * w = μ n*l*IB = μ n I

*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.***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 %.

- for axially symmetric currents,
- We define the
**magnetic flux**:Φ

Because there are no point-like objects with magnetic "charge", Gauss' law gives us_{B}≡ ∫**B ⋅ dA**.∫

_{closed surface}dΦ_{B}= 0. - 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**:∇ ⋅

and the**A**≡ dA_{x}/dx + dA_{y}/dy + dA_{z}/dz,**curl**:∇ ⊗

With these definitions,**A**≡ (dA_{y}/dz - dA_{z}/dy, dA_{z}/dx - dA_{x}/dz, dA_{x}/dy - dA_{y}/dx).**B**= ∇ ⊗**A**.∇ ⋅

and is then a result of the identity**B**= 0,∇ ⋅ (∇ ⊗

**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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