. A text-book of electrical engineering;. e •(38). T ^I = —: tan a 2,ir If the horizontal component of the earths field at any place is known,the strength of current can thus be calculated from the values of r and , a standardised tangent galvanometer can be used to deter-mine the horizontal component. 58 Electrical Engineering We wi]l now consider the case in which the pole m lies on the axis A Aof the coil, but at a distance from it (Fig. 49). If the line a joining the poleto the circumference of the coil makes an angle a with the axis, then r sin a Now the element of condu
. A text-book of electrical engineering;. e •(38). T ^I = —: tan a 2,ir If the horizontal component of the earths field at any place is known,the strength of current can thus be calculated from the values of r and , a standardised tangent galvanometer can be used to deter-mine the horizontal component. 58 Electrical Engineering We wi]l now consider the case in which the pole m lies on the axis A Aof the coil, but at a distance from it (Fig. 49). If the line a joining the poleto the circumference of the coil makes an angle a with the axis, then r sin a Now the element of conductor at B is perpendicular to the plane of thepaper and is therefore at right angles to the line a which lies in that sin 0 = I in equation (30) on page 54, we get ,j. .dl .dl.*/=—^2—= . sm^ a This force acts perpendicularly to the plane containing the pole and theelement of conductor, that is, in the direction CD. The component actingalong the axis is CE = df. sin a. Adding up the forces due to all the ele-. Fig. 49 ments of conductor constituting the coil, we find that all the componentssuch as ED, at right angles to the axis, neutralise one another, while thecomponents along the axis must be added. The total force acting on thepole m is therefore along the axis A A and has a magnitude / = 5^/. sm a = 2 . ^l. Since Yidl is equal to 277/, this reduces to , 2Trm .1 . .J = . sm** a ■(39)- To find the field strength at point C, that is, the force on a unit pole atthis point, we must put wi = i in equation (39), giving 17 H. = . sm-* a .(40). This equation will prove of great assistance to us in the next section incalculating the field of a solenoid. 27. Magnetic Field of a Solenoid 59 27. Magnetic Field of a Solenoid. The lines of force due to two parallel conductors carrying currents in thesame direction combine to form lines of force which encircle both con-ductors. Two conductors are shown, for example, in Fig. 50, passing ver-tically thro
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