Maxwell's Equations:

Note to the student: This section is reserved for advanced students, with background in electricity and magnetism, and vector differential equations.

Problem:
Given Maxwell's four equations, demonstrate the existence of a vector magnetic potential and a scalar electric potential. Derive field equations for these potentials.

Solution:
Maxwell's four equations read:

x H =

₀

t + j

x E = -

₀

^. H = 0

^. E =

₀

where H is the magnetic field (A/m), E is the electric field (V/m), j is the vector current density (A/m²), ₀ = 8.8542 x 10^-12 F/m is the permittivity of free space, ₀ = 4 x 10^-7 H/m is the permeability of free space, and is the scalar charge density (C/m³).

Existence Of The Vector And Scalar Potentials

Let us begin with eq. 3. By a theorem of vector calculus, ^. H = 0 if and only if there exists a vector field A such that

H =

x A

Call A the vector magnetic potential.

Let us now take eq. 2. We will find a characteristic solution E_c and a particular solution E_p. The total electric field will then be the sum E_c + E_p.

a. Characteristic solution: x E_c = 0. By another theorem of vector calculus, this expression can be true if and only if there exists a scalar field such that E_c = - . Call the scalar electric potential.

b. Particular solution: x E_p = - ₀ H/t, and H = x A allows us to write

x E_p = - ₀ ( x A)/t = x (- ₀ A/t)

We may obtain a particular solution E_p by simply choosing E_p = - ₀ A/t.

c. Total electric field:

E = E_c + E_p = - - ₀A/t

6.

Relationships Between The Vector And Scalar Potentials

Let us now substitute the expressions derived above into eqs. 4 and 1. From eq. 4, we obtain

^.(-

₀

t) =

₀

and from eq. 1, we obtain

x (

x A) =

₀

t)/

t + j

By still another theorem of vector calculus, we have the identity

x (

x A) =

(

^.A) -

²A

so that eq. 8 becomes

(

^. A) -

²A =

₀

t)/

t + j

8a.

or, after some simplification

(

^. A) -

²A = -

(

₀

t) -

₀

²A/

t² + j 9.

Since x grad = 0, eq. 9 may be manipulated by taking the curl of both sides. The terms involving the gradient then vanish leaving us with the identity

x (

²A) =

x (

₀

²A/

t² + j)

from which we may infer, without loss of generality, that

²A =

₀

²A/

t² + j

10.

²A -

₀

²A/

t² = j

10a.

Eq. 10a is one of the field equations we sought. We may simplify eq. 10a somewhat if we recognize that ₀ ₀ = 1/c² where c is the speed of light. We now introduce a new operator ² defined by

² =

² - (1/c²)

² /

t²

so that eq. 10a becomes

²A = j

11.

Using eq. 10 in eq. 9, we obtain

(

^. A) = -

(

₀

from which we may infer, again without loss of generality, that

^.A = -

₀

12.

We must next rewrite eq. 7 by distributing the operator .

^. (-

₀

t) =

₀

becomes

₀

(

^.A)/

t =

₀

7a.

Substituting eq. 12 into eq. 7a gives

₀

t)/

t =

₀

which may be simplified:

₀

t² = -

₀

13.

or, recalling that ₀ ₀ = 1/c², and using the operator ²

= -

₀

13a.

Eq. 13a is the other field equation that we sought.

To summarize:

The existence of a vector magnetic potential A and a scalar electric potential was demonstrated. The respective field equations for A and were found to be ²A = j
and ² = - /₀

where j is the vector current density (A/m²), is the scalar charge potential (C/m³), and ₀ = 8.8542 x 10^-12 F/m is the permittivity of free space.
Dimensional analysis on the above equations shows that A has units of electric current (amperes) and has units of electric potential (volts).

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