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EXAMPLE A nonmagnetic rotor containing a single-turn coil is placed in a uniform magnetic field of magnitude Bo, as shown in Fig. 3.2. The coil sides are at radius R and the wire carries current J Uniform magnetic field, Boj Wire 2, current / |_| out of paper Figure 3.2 Single-coil rotor for Example 3.1. 3.4 Forces and Torques in Magnetic Field Systems as indicated. Find the 6-directed torque as a function of rotor position a when J = 10 A, By = 0.02 T and R = 0.05 m. Assume that the rotor is of length / = 0.3 m. & Solution ‘The force per unit length on a wire carrying current / can be found by multiplying Eq. 3.6 by the cross-sectional area of the wire. When we recognize that the product of the cross-sectional area and the current density is simply the current I, the force per unit length acting on the wire is given by F=IxB Thus, for wire | carrying current / into the paper, the 6-directed force is given by Fi, = —1 Bol sina and for wire 2 (which carries current in the opposite direction and is located 180° away from wire 1) Fy, = —1 Bol sina where / is the length of the rotor. The torque T acting on the rotor is given by the sum of the force-moment-arm products for each wire T = —21 ByRI sina = —2(10)(0.02)(0.05)(0.3) sina = —0.006sina N-m Wire 1, current 115 
