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were in amperes, it would be the “ absolute" calibration curve of the instrument.

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2. Delineation of Lines of Magnetic Force

(Compass-Needle Method). Introduction. In the medium surrounding any magnetic body whatsoever a "magnetic field" of force always exists, and the direction in which a free north pole would move when acted on by such a field is termed a " line of magnetic force." Now, the magnetic field in the vicinity of, for instance, a bar magnet consists of a large number of " lines of force," each forming a closed curve or loop, and the intensity of such a field will be represented by the number of lines emerging from one-half of the magnet which pass through the surrounding air and enter the other half of the magnet, thus completing their paths through the magnet. Thus, a line of force denotes the direction of the force acting on a north pole placed in the field, and to find this direction, i.e. that in

which a north pole would move if free to do so, we may proceed as follows, assuming that the “law of inverse squares" holds good, namely, that the force exerted between two magnetic poles, whether of attraction or repulsion, is inversely proportional to the square of the distance between them. Then, referring to Fig. 3, the force

FIG. 3.


and may


acting on the free north pole at P due to N is of

(NP)? be taken as Pn, direction of which will be along NP produced. Now, S exerts on the north pole at P an attractive force of


(NP)2 which must be denoted by a line = Pn.

= Ps.


(SP) the free pole at P being acted on by two forces, Pn and Ps, will have a resultant force represented, in magnitude and direction, by the diagonal PQ of the parallelogram, of which Pn and Ps are adjacent sides, and so on for the rest. Thus PQ, QR, RT .. form elements of the line of force, which will eventually reach the south pole of the magnet. We cannot, however, separate a north pole from its conjugate south pole in practice; but, notwithstanding this, a short compass needle will take up positions represented by those elements of the line of force with which its magnetic axis will coincide, and hence enable the whole line to be mapped out. It must be clearly understood that the earth's field will modify the distribution of the field due to the magnet, as also would any other field, and the object of the present experiment is to determine the distribution of the lines of force when so acted upon.

Apparatus.—Short bar magnet, small compass needle, large sheet of drawing-paper, and drawing-pins.

Observations.-(1) Pin the drawing-paper to the table or a drawing-board, with one of its sides parallel to the magnetic meridian of the earth.

(2) Place the bar magnet near one side of the paper with its magnetic axis parallel to the meridian, and its N. pole pointing towards the N. geographical pole of the earth, and make a pencil line round it at a distance of say about i cm.

N.B.—The magnetic N. pole of the earth, though not actually coincident, is practically the same as the S. geographical pole.

(3) Divide the long sides of the line into about six equal parts, and the end or short lines into about four equal parts, and place the compass so that one end (call it A) is just opposite or over one of the marks, o. Then make a pencil point or mark, 1, just opposite the other end (call it B) of the needle.

(4) Next slide the compass along in the direction it points until end A is over mark 1; then make another fresh mark, 2, just opposite end B, and so on until the whole curve o 1 2 3 4 is mapped out. Then draw a neat curve through all these points.

(5) Repeat (2)-(4), starting from each of the other marks on the rectangular line.

(6) Repeat (2)-(5) with the magnet placed near the other side of the paper, but reversed so that its S. pole now points toward the N. geographical pole of the earth.

Inferences. What can you infer from the respective distributions of the lines ? Point out their bearing on fundamental principles involved with magnetic poles.

3. Distribution of Magnetism in Bar Magnets

(Method of Oscillations). Introduction. From the remarks and results of observations in the last experiment, we may gather that every uniformly

magnetized magnet possesses a region, approximately N

midway between its ends, from which no lines of force emerge into the air, i.e. a region at which there is no free magnetism, which is called the equator of the magnet. The intensity of magnetization or the number of lines of force inside the magnet is greatest here at this point, and diminishes towards the ends, where the amount of free magnetism is a maximum. Hence any

external magnetic effect of the magnet on neighbouring E

bodies will be a minimum at the equator and a maximum Hc at the poles. Now, suppose a short magnetic needle,

ns, is suspended in a frame of brass, E, which can slide

along a bar magnet, NS, and be clamped in any part S

by a screw, c. Then from p. 221, we see that if the

needle oscillates in front of NS, which is placed vertically so that the magnetic axis of ns cuts it, and is also parallel to the magnetic meridian, thenK

47'K T = 27

or M(Hm+ H) M(H + Hg)

T? where T

= time of one complete oscillation in seconds of the needle having a moment of inertia K and magnetic moment M, He being the horizontal intensity of the earth's field, and H,n being the horizontal component of the magnet's total field; for manifestly the earth's field is superimposed on that of NS, resulting in a total field (H + Hg) acting on ns. The + and sign denotes whether the two fields are acting summationally or differentially respectively on the needle, which will depend on what half of NS the needle is opposite. If the north magnetic pole of the earth is on the left, and the south magnetic pole of the earth on the right of Fig. 4, then, when the needle is below the equator of NS, the north pole of it will point as shown, and we evidently have a differential effect of the fields (H... - Hz), and the south magnetism

FIG. 4.

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of NS will preponderate over the north magnetism of the earth. If the needle is above the equator, we have a summational effect of the fields (H2 + He). If ni number of “transits” of the needle per second, i.e. the number of times one end swings across the magnetic meridian per second, T, = and the above

N, formula becomes

4πKn, ?

TK (H. + Hg) =

-.1', or (H + H) on?

M If ns is very close to the magnet, H,n will be proportional to the amount of free magnetism at any point, and He will be equal to TK

-n, where n = number of oscillations per second in the earth's M M field alone. Then, for two different positions of ns on the magnet on the same side of its equator, we have for summational

72K effect of magnet and earth on the needle, Hml (11,2 - 11%), and for summational effect of magnet and earth on the needle,

2K Hn2 (n,? no), where n, and n, = number of transits per

M second in the two positions. Hence

ni? 112

11, 112 For the differential effect

H, n2 + n2

11," n In other words, the strengths of two single fields are proportional to the squares of the number of oscillations performed in each by the same suspended needle.

Thus, by placing the needle at different parts of the magnet, and obtaining the number of vibrations in the same time at each, the squares of these will be proportional to the free magnetism at each.

Observations.-(1) Remove the magnet from its supporting clamp (not shown), and take off the clamp E carefully. Then, bringing the magnet up to a compass card or other galvanometer needle, note the effect, and thus determine which is the north pole.

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