Part 1: Basic Magnetostatics

The basic law of electrostatics is Coulomb’s Law. When working with charge densities rather than discrete points, the more useful form of this is Gauss’ Law (see here and here for previous uses). In integral form, Gauss’ Law reads:

.

Using the divergence theorem, this can be converted into the differential form

.

Comparably, the absense of magnetic monopoles gives the analogous equation

.

Extending beyond electrostatics into magnetostatics, we begin with the conservation of charge in relation to current. The continuity equation is

(as mentioned here). For a steady-state magnetic system, we need no change in the net charge density anywhere, and thus .

For a steady current in an infinitely thin wire, the (static) magnetic field is described by the Biot-Savart Law

,

where *I* is the current, *d***l** is the differential element vector of the wire oriented in the direction of the current, is the displacement unit vector pointing from the wire element to the point where we are measuring the field, and *r* is the distance from the wire element to the point where the field is computed; the integral is computed over the wire, which is either a loop, or else extends to infinity (no endpoints). Using the knowledge that the displacement vector is , then we can rewrite it as

.

Let us compute the magnetic field from an infinitely long, straight wire with current *I*. We let our wire be the *z* axis, and let *z*‘ be the coordinate of our current element, and let us choose our *z*=0 plane to be the plane containing the point we want to measure our field, which due to symmetry should be independent of *z*, and dependent only on the distance from the wire. Then , , and so

,

which in cylindrical coordinates is . Thus, we have

.

Extending the Biot-Savart law from an infinitely thin wire to a non-zero thickness, we replace the current differential element with , where **J** is the current density, and the integral is over space; using **r’** for the position vector of current elements and **r** as the position vector for the location at which we compute the field, we obtain

Viewing as a function of **r** (inverse-square vector field from the point **r’**), we can use that it is the negative gradient of the inverse distance scalar:

,

where the gradient involves **r**, not **r’**. Now, for any vector field **a** and scalar field *ψ*, it is an identity that

,

thus

,

and so

.

Now, since our curl is with respect to **r**, and the current density is a function of **r’** only, , and so

, and since the integral is over **r’**, it will commute with the above curl over **r**, so we obtain

.

Note that since the divergence of a curl is always zero, this automatically gives us .

Next, let us take the curl of our magnetic field:

.

Here, we need to use the identity for curl of a curl: for any vector field **a**,

,

giving us

.

Now, the laplacian of the scalar field is equal to the divergence of it’s gradient, which is an inverse square vector field, and so is a three-dimensional Dirac delta:

, meaning that the rightmost integral term becomes

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We can also use the fact that

,

where indicates the gradient as a function of **r’**; this gives us

;

this latter integral may be transformed via vector integration by parts; since

, and since our integral being over all space means the bounding surface term vanishes, we obtain

,

so

.

Now, recall that for a magnetic steady-state, we needed , and so the numerator of the integrand in the above is zero, giving us . This is the differential form of Ampère’s law. The corresponding integral form may be obtained from this (or vice-versa) via Stokes’ theorem, and says

,

or in words, the line integral of the magnetic field over a closed curve is equal to the total current *I* passing through the curve.

Tags: Continuity Equation, Curl, Electricity & Magnetism, Electrostatics, Friday Physics, Gauss' Law, Integration by Parts, Magnetostatics, physics

June 18, 2010 at 12:21 am |

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July 16, 2010 at 2:55 am |

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July 23, 2010 at 12:21 am |

[…] and so the sum over all of these elemental loops becomes a volume integral: . Now, recall that Ampère’s Law states that , and so, using this to replace the current density in the above, we see . Now, the […]

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