Electrical measurementsA laboratory manual . ^m XwiJs Fig. 132. be toward the west; the force at 0 due to it will be m^ (• - ihy , directed toward the east. The force due to MAGNETISM. 293 mi ^, directed toward the the positive pole will be west. The total force on unit pole will then be cf=mi \(:r-¥d (r+ihY) 2m,l, X 4r^) 2gj(3r 1 + ?T , 3^ _A_ 4- etc. directed tow^ard the east. Neglecting higher powers thanthe second in the expansion, this may be written 23m 8f= ^(-j> (8) Second Method. In this methodlet the positive pole of AB () be toward the east. Thenthe force on a unit negative
Electrical measurementsA laboratory manual . ^m XwiJs Fig. 132. be toward the west; the force at 0 due to it will be m^ (• - ihy , directed toward the east. The force due to MAGNETISM. 293 mi ^, directed toward the the positive pole will be west. The total force on unit pole will then be cf=mi \(:r-¥d (r+ihY) 2m,l, X 4r^) 2gj(3r 1 + ?T , 3^ _A_ 4- etc. directed tow^ard the east. Neglecting higher powers thanthe second in the expansion, this may be written 23m 8f= ^(-j> (8) Second Method. In this methodlet the positive pole of AB () be toward the east. Thenthe force on a unit negative pole ^^at a distance r north from themiddle of AB^ due as before tothe negative pole of AB, will benil (2 , Zf \ directed from B. The force due to the positive pole willbe the same in magnitude, andwill be directed toward A, Forconvenience in drawing Fig. 133,it has been assumed that the polesare at the ends of AB. In realitythey should be further back. Re-solving these forces into south andeast and north and east compo-. Fig. 133. 294 ELECTRICAL MEASUREMENTS. nents, we fiiid that the north and south componentsannul one another, and the east components produce aforce on unit negative pole, directed toward the east. As before, this may be writtenwithout sensible error, =?(-1) (») Returning now to the first method, we may suppose ashort magnet ns (Fig. 132) of length I and pole strengthm suspended at 0. Call the deflection produced byAB </). Then for equilibrium the moments of the twocouples acting on ns must be equal, or &€ml sin (j) = Efml cos =^^{l+-^. . (11) Eliminating c from (10) and (11) we obtain-r^nan<^^ In a similar way for the second method we findequilibrium of the moments of the two couples when a8tan^=f (l-£;), . (12) MAGNETISM. 296 and g<gtan — r^ tan ^ Correction for Induced Magn
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