Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/247

rectangle, are y₀ and z − ½dz. The corresponding value of G is

$G=G_{0}-{\frac {1}{2}}{\frac {dG}{dz}}dz+\And c.,$

(8)

and the part of the value of p which arises from the side A is approximately

$G_{0}dy-{\frac {1}{2}}{\frac {dG}{dz}}dydz.$

(9)

Similarly, for B,

$H_{0}dz+{\frac {1}{2}}{\frac {dH}{dy}}dydz.$

For C,

$-G_{0}dy-{\frac {1}{2}}{\frac {dG}{dz}}dydz.$

For D,

$-H_{0}dz+{\frac {1}{2}}{\frac {dH}{dy}}dydz.$

Adding these four quantities, we find the value of p for the rectangle

$p=\left({\frac {dH}{dy}}-{\frac {dG}{dz}}\right)dydz.$

(10)

If we now assume three new quantities, a, b, c, such that

${\begin{aligned}a&={\frac {dH}{dy}}-{\frac {dG}{dz}},\\b&={\frac {dF}{dz}}-{\frac {dH}{dx}},\\c&={\frac {dG}{dx}}-{\frac {dF}{dy}},\end{aligned}}$

(A)

and consider these as the constituents of a new vector ${\mathfrak {B}}$, then, by Theorem IV, Art. 24, we may express the line-integral of ${\mathfrak {A}}$, round any circuit in the form of the surface-integral of ${\mathfrak {B}}$, over a surface bounded by the circuit, thus

$p=\int {\left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}\right)ds}=\iint {(la+mb+nc)dS},$

(11)

or

$p=\int {T{\mathfrak {A}}\cos \epsilon \,ds}=\iint {T{\mathfrak {B}}\cos \eta \,dS},$

(12)

where ε is the angle between

${\mathfrak {A}}$ and ds, and η that between ${\mathfrak {B}}$ and the normal to dS, whose direction-cosines are l, m, n, and $T{\mathfrak {A}}$, $T{\mathfrak {B}}$ denote the numerical values of ${\mathfrak {A}}$ and ${\mathfrak {B}}$

Comparing this result with equation (3), it is evident that the quantity I in that equation is equal to ${\mathfrak {B}}\cos \eta$, or the resolved part of ${\mathfrak {B}}$ normal to dS.

592.] We have already seen (Arts. 490, 541) that, according to Faraday s theory, the phenomena of electromagnetic force and