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CHAPTER VIII

MAGNETISM

Note.--All quantities are expressed in terms of the C.G.S. units. For the definitions of magnetic units and their dimen. sions, see pp. 4 and 15.

1. A magnetic pole of strength 90 is found to attract another pole 2 cm. from it with a force equal to the weight of a gramme : what is the strength of the second pole ? By the law of inverse squares, the force exerted between the two poles is equal to the product of their strengths divided by the square of their distance apart. This is equal to 981 dynes (the weight of a gramme), so that if P be the strength of the second pole,

981=P x 90/22 and

P=4 x 981/90= 43.6. 2. The strength of a certain magnet-pole is 27: find the intensity of the magnetic field 3 cm. away from it, assuming the magnet to be so long that the influence of the other pole may be neglected. What force would be exerted by it upon a pole of strength 32 at a distance of 12 cm. ? The intensity of the magnetic field at any point is measured

by the force experienced by a unit magnetic pole placed at the point. The force exerted by the given magnet-pole on an unit pole 3 cm. away from it is 27 x 1/32= 3 dynes ; and hence the strength of the field at this distance is 3.

In the second case the force would be 27 x 32/(122)=6

dynes. 3. What is the force exerted between two poles of strength 32 and 36 units at a distance of 12 cm. from one another?

4. The repulsive force between two poles is 20 dynes when they are 4 cm. apart : what will it be when the distance between them is increased by i cm. ?

5. A magnet - pole of strength to attracts another pole 5 cm. from it with a force of 2 dynes : what is the strength of the second pole?

6. The distance between two equal magnet-poles is 8 cm., and they repel one another with a force of 5 dynes : find the strength of each.

7. A magnet 8 cm. in length lies in a field of intensity H=0:18, and the strength of each of its poles is 5. Find the moment of the couple required to deflect it (1) through an angle of 30° from the magnetic meridian, (2) at right angles to the magnetic meridian. The force acting on each pole in both cases is mH = 5 x 0.18

=0.9. (1) The arm of the couple, or perpendicular distance between

the forces acting on the two poles, is 1 sin d = 8 sin 30o = 8/2=4, and the moment of the couple is m Hlsin o=o.9

X4=3.6. (2) When the needle is at right angles to the meridian the

arm of the couple is equal to the length of the needle, and

the moment of the couple is 0.9 x8=7.2. 8. Given that the dimensions of strength of pole are M LÍT, show that the dimensions of strength of field are MłL-T. What is the strength of a pole which is urged with a force of 9 dynes when placed in a field of intensity 0.5?

9. A freely suspended magnetic needle is deflected (1) through an angle of 45°, (2) through an angle of 60 from the magnetic meridian. Compare the couples which tend to bring the needle back to its position of rest in the two cases.

10. Prove that the magnetic force (or intensity of field) due to a bar magnet at any point on the axis of the magnet produced is equal to 2 Md/(a2 - 12), where M is the magnetic moment of the magnet, 2l its length, and d the distance of the point from its centre. Find the force due to each pole separately and subtract.

Observe that since the length of the bar magnet is 2l, its

magnetic moment is M = 2ml. 11. Prove that the magnetic force due to the same magnet at a point opposite to its centre and at a distance d from it is equal to M/(a12 + 12), and acts parallel to the axis of the magnet. The forces due to the two poles are equal, and their com

ponents perpendicular to the axis are equal and opposite.

Find the components parallel to the axis and add. 12. A magnetic needle is deflected through an angle a from the meridian by a bar magnet of magnetic moment M placed "broadside M on” (i.e. in the position shown in Fig. 3, the line joining the centres of the two magnets being in the meridian MR, and the axis of the bar-magnet perpendicular to it.) Prove that if H be the strength of the field, and d the distance between the centres of the magnets, M/H = (d2 +12) tan a. Find the moments of the couples due to the

action of the bar-magnet and of the earth's

field, and equate. 13. Prove that the equation of equilibrium in the preceding example would take the form

M/H = (dl2 – 12)? tan a/2d, if the bar-magnet were placed “end on," i.e. in an east

'S

n

Fig. 3.

and west position, with the centre of the deflected magnet lying on the axis of the bar-magnet produced.

Note on the Time of Vibration of a Magnet.The small oscillations of a magnet'are governed by laws similar to those which regulate the vibrations of a pendulum swinging through a small arc.

The time of a complete vibration of a simple pendulum is t=27 Nig: the time of a complete vibration of a magnetic needle suspended horizontally is t= 27 VI/K, where I is the moment of inertia of the needle, and K is the moment of the directive force tending to restore the needle to its position of rest. The quantity I depends upon the mass and dimensions of the needle ; K is the product of two factors, M and H, M being the magnetic moment of the magnet, and H the horizontal intensity of the earth's magnetic force. The force acting on a needle is therefore proportional to the square of the number of oscillations which it makes in a given time.

14. A magnetic needle is suspended horizontally at a considerable height above a bar-magnet which lies on the floor underneath it. When the north pole of the bar-magnet points southward, the needle makes 14 vibrations per minute ; and when its north pole points northwards, the needle makes 8 vibrations per minute. At what rate would the needle vibrate under the action of the earth's force above ?

Let n denote the number of oscillations made per minute by

the needle under the action of the earth's force H. Then

H=kn”, where k is a constant. In the first position of the bar-magnet the needle is vibrating

in a field the strength of which is H +F, where F is the force due to the bar-magnet. In the second position of the bar-magnet its action is opposed to that of the earth,

and the resultant field is of strength H - F. Thus

H+F= kx (14)2 = 196 k, and

H - F=kx (8)2 = 64k.

H=130k.
Again, since H=kn2, it follows that n= V130= 11.4.

H

Fig. 4.

15. A compass needle makes 50 oscillations per minute at a place where the dip is 64°, and 48 oscillations per minute at another place where the dip is 71°. Compare the value of the total magnetic force at the two places. Let I denote the total force which acts along the line of dip,

H the horizontal component, and 0 the
angle of dip; then H/I = cos 0, or
I=H/ cos @ (Fig. 4). Thus at the
first place we have I = H/ cos 64° =
H/0.438, and at the second place I'=

H' cos 71° = H/0.326.
The vibrations of the compass-needle are

controlled by the horizontal component
of the total force, therefore

H:H'=(50)2 :(48)2 = 2500 : 2304,
and
I:I'= (2500/0.438):(2304/0.326)

=I:1.238. 16. A magnetic needle makes 100 oscillations in 8 min. 20 sec. under the action of the earth's force alone. Under the combined action of the earth and a magnet A it makes 100 oscillations in 7 min. 30 sec., and when the magnet A is replaced by another B the needle makes 100 oscillations in 6 min. 40 sec. Compare the magnetic moments of A and B.

17. A compass needle makes 10 oscillations per minute under the influence of the earth's magnetism alone. When the north pole of a long magnet A is held i ft, south of it, the needle makes 12 oscillations per minute ; and when the north pole of another magnet B is held in the same position, the number of oscillations per minute increases to 15. Compare the pole-strengths of the magnets A and B.

18. At Berlin the total magnetic intensity is 0.48 (in C.G.S. units) and the dip is 64°: at New York the total intensity is 0.61 and the dip 72°. If a magnet

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