The electromotive force will be the product of the velocity of travelling, the distance between the rails, and the vertical intensity, that is,

(44.7 × 30) (254 × 56·5) (438) = 84300

8. Find the electromotive force at the instant of passing the magnetic meridian, in a circular coil consisting of 300 turns of wire, revolving at the rate of 10 revolutions per second about a vertical diameter; the diameter of the coil being 30 centims., and the horizontal intensity of terrestrial magnetism being 1794, no other magnetic influence being supposed present.

The numerical value of the lines of force which go through the coil when inclined at an angle to the meridian, is the horizontal intensity multiplied by the area of the coil and by sin ; say nHa2 sin 0, where H = ∙1794, a = 15, and n = 300. The electromotive force at any instant is the rate at which this quantity increases or diminishes; that is, nНπа2 cos 0 . w, if w denote the angular velocity. At the instant of passing the meridian cos is 1, and the electromotive force is «Ηπαω. With 10 revolutions per second the value of o is 27 × 10.

Hence the electromotive force is

*1794 × (3*142)2 × 225 × 20 × 300

This is about I of a volt.

42

=

2'39 × 106.

190. To investigate the magnitudes of units of length,

mass, and time which will fulfil the three following conditions:-

1. The acceleration due to the attraction of unit mass

at unit distance shall be unity.

2. The electrostatic units shall be equal to the electromagnetic units.

3. The density of water at 4° C. shall be unity.

Let the 3 units required be equal respectively to L centims., M grammes, and T seconds.

We have in C.G.S. measure, for the acceleration due to attraction (§ 72),

This equation expresses the first of the three conditions.

Substituting L3 for M in the first equation, we find

Introducing the actual values of C and 7, we have approximately

T= 3928, L 1'178 x 10", M = 163 × 102;

that is to say,

=

The new unit of time will be about 1h 51m;

The new unit of length will be about 118 thousand earth quadrants;

The new unit of mass will be about 2.66 × 1014 times the earth's mass.

Electrodynamics.

191. Ampère's formula for the repulsion between two elements of currents, when expressed in electromagnetic units, is

cc ds. ds'

where c,

(2 sin a sin a' cos 0 cos a cos a'),

denote the strengths of the two currents;

ds, ds the lengths of the two elements;

a, a' the angles which the elements make with the line joining them;

r the length of this joining line;

the angle between the plane of r, ds, and

the plane of r, ds.

For two parallel currents, one of which is of infinite

length, and the other of length 7, the formula gives by integration an attraction or repulsion,

21

where D denotes the perpendicular distance between the

currents.

Example.

Find the attraction between two parallel wires a metre long and a centim. apart when a current of is passing through each.

I
IO

Here the attraction will be sensibly the same as it one of the wires were indefinitely increased in length, and will be

that is, each wire will be attracted or repelled with a force of 2 dynes, according as the directions of the currents are the same or opposite.