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

592.]
215
MAGNETIC INDUCTION.

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

 $G_0 dy - \frac{1}{2} \frac{dG}{dz}dy dz.$ (9)
 Similarly, for B, $H_0 dz + \frac{1}{2} \frac{dH}{dy}dy dz.$
 For C, $-G_0 dy - \frac{1}{2} \frac{dG}{dz}dy dz.$
 For D, $-H_0 dz + \frac{1}{2} \frac{dH}{dy}dy dz.$

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{align} a &= \frac{dH}{dy} - \frac{dG}{dz}, \\ b &= \frac{dF}{dz} - \frac{dH}{dx}, \\ c &= \frac{dG}{dx} - \frac{dF}{dy}, \end{align} (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