By P. Hammond

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Example text

The force P = 3i + 5j moves its point of application from (1, 2) to (4, 3). What is the work done? 4. Show that the area of a parallelogram of sides a and b is given by A = a x b. What is meant by the direction of an area? 5. Show that A x (B x C) = В (A . C) - С (A. B). 6. Explain what is meant by the gradient of a scalar quantity. Express the gradient of the scalar potential V in cylindrical (r, 0, z) and spherical (r, 0, φ) coordinates. 7. The vector г points from the source located at (JC', y', z') to the field point (x,y, z).

Knowledge of the total source can be obtained from the field sur­ rounding the body. Gauss's theorem, for example, tells us that we can find the total enclosed charge from thefluxthrough the enclosing surface, and, similarly, we can find the total enclosed current from the circulation of the magneticfieldstrength around the perimeter [eqn. 26)]. Better still, we shall show in the next chapter, § 4, that we can obtain not only the magnitude of the source but also the force on it if we know the field strength at every point on an enclosing surface.

The line integral is § H . 28) ду J The line integral has been taken in the counter-clockwise direction and with a right-handed set of axes this associates the integral with the posi­ tive z-direction. The current flow through the small loop is Jz òx ôy. Thus õHy дНх = / , . 31) Thus the small line integrals are associated with a vector which we shall call curl H, where i j к curl H = õ õ Ô = J дх ду dz Нх Ну Ηζ and by comparison with eqn. 4) we can write this in terms of the operator V curl H = V x H = J.