Magnetic Moments of Magnets. Deflection Method. Solution of Inferences.-A. Position.-Let the horizontal intensity H of the earth's field act parallel to the needle ns in its position of rest. Also let the force F of the bar magnet, whose length = = 27, on one of the poles of the needle be always at right angles to this position of rest; therefore F m1H tan on either pole of the needle. Now, by the law of "inverse squares," the force exerted between two magnetic poles of strengths m, m2 mm2 = da dynes, where d = distance between them in centimetres, which force is a mutual attraction if the poles are of opposite kinds, and a repulsion otherwise; Therefore force exerted by N pole on one pole of the needle = mm (d − 1)2 where m = strength of magnet's poles, and m, that of the needle's. But these forces act in opposite directions on the same pole of ns. Hence resultant force F of the magnet on either pole of the mm1 mm14dl = m1H tan 0. mm1 needle = = (d2 - 12)2 Assumptions.-(1) That the poles of the magnet are at the ends, and therefore at a distance 2/ apart. (2) That the length 27 is very large compared with the length of needle, so that the mutual forces between the magnet are all parallel. (3) That the forces are unaltered in magnitude by the deflection of the needle. Deflection Method. Solution of Inferences.-B. Position.-Let the horizontal in But these combine to an equivalent single F acting parallel to NS, which pulls the s pole of the needle counter clockwise. which is the same force acting in the same direction on n. and couple acting on needle = m1H2l, sin @ due to earth's magnetism; and when the deflection is steady, these couples just balance. where m2l = M, the magnetic moment of the magnet. Vibration Method. Solution of Formula.-Let the magnet be deflected from its position of rest in the magnetic meridian, through a small angle • ; then the horizontal component H of the earth's field exerts a torque, tending to bring the magnet back to its original position of rest. Now, let m = the strength of each of the magnet's poles, and 2/ the distance between them, and K = = moment of inertia about the axis of vibration. Then, since a magnet, whose length is large compared with its breadth, may be assumed to have two poles of equal strength, but of opposite kinds, situated at its ends The magnetic moment M = 2ml and the force acting on each pole therefore the couple exerted by these = mH two equal and opposite forces} = mH x 27 sin • = MH sin @ Now, for translational motion we have force acting = mass × acceleration; and for rotational motion we have couple acting = moment of inertia x angular acceleration ; whence MH sin @ = Κω where & = angular acceleration. But the equation of motion for a simple pendulum is g sin @ = where /= length of pendulum, and g the acceleration due to gravity. Thus the periodic time of vibration T = 2π/ consequently, for the magnet T = 2π/ K MH on the assumption that @ is small, say, not more than 5° or 10°, and that the motion is one of pure rotation. Measurement of Resistance. Substitution Method. Solution of Inferences.-Let g = resistance of the galvanometer, and R, that of the known and unknown; also let E = E.M.F. of the battery, and its internal resistance. Then, if C is the current flowing both when R and are separately in circuit, as is indicated by the same galvanometer deflection in each case, we have, by Ohm's law— E E C=R+b+8=r+6 +8 Hence R = r (the unknown resistance). This is based on the assumption that both E and b are constant throughout any pair of readings, which may not be true unless the cell has a fairly constant E.M.F. (such as that of a Daniell), and is allowed to send only a feeble current by having the external resistance as high as possible. Wheatstone Bridge Method. Referring to Fig. 19 (p. 37), let V1, V2, V3 be the potentials of the points A, H, and D respectively; then, when no current flows through the galvanometer, i.e. when the bridge is balanced, and there is therefore no deflection, V, will be the potential of the point C also. Hence, if C1, C2, C3, C, are the currents flowing through the resistances 71, 72, 73, respectively of the "arms," then, since no current passes through G, we have, by Ohm's law 2 |