Electromagnetism 153 Worked Example 5.8 Q A pair of pole pieces 5 cm by 3 cm produce a fl ux of 2.5 mWb. A conductor is placed in this fi eldwith its length parallel to the longer dimension of the poles. When a current is passed throughthe conductor, a force of 1.25 N is exerted on it. Determine the value of the current. If the conductor was placed at 45° to the fi eld, what then would be the force exerted? A 2.5 10 3 Wb; ℓ 0.05 m; d 0.03 m; F 1.25 N csa of the field, mflux density, ABA00500350 32...11tteslaTand since , then sin 250506679033...11111111FBFBIII sin newton, so amptherefore ...256670 05515 A Ans FBFI sin newton, where so henc4566750 050 70711...ee, N F0 884.Ans The principle of a force exerted on a current carrying conductor as described above forms the basis of operation of a linear motor. However, since most electric motors are rotating machines, the above system must be modiﬁ ed. 5.6 The Motor Principle Once more, consider the conductor formed into the shape of a rectangular loop, placed between two poles, and current passed through it. A cross-sectional view of this arrangement, together with the ﬂ ux patterns produced is shown in Fig. 5.14 . The ﬂ ux patterns for the two sides of the loop will be in opposite directions because of the direction of current ﬂ ow through it. The result is that the main ﬂ ux from the poles is twisted as shown in Fig. 5.15 . This produces forces on the two sides of the loop in opposite directions. Thus there will be a turning moment exerted on the loop, in a counterclockwise direction. The distance from the axle (the pivotal point) is r metre, so the torque exerted on each side of the loop is given by TFrFBI newton metrebut sin newton, and sin 1