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Unformatted text preview: Solutions, Assignment 20 27.42. I DENTIFY : sin IAB = . The magnetic moment of the loop is IA = . S ET U P : Since the plane of the loop is parallel to the field, the field is perpendicular to the normal to the loop and 90 = . E XECUTE : (a) 3 (6.2 A)(0.050 m)(0.080 m)(0.19 T) 4.7 10 N m IAB  = = = (b) 2 (6.2 A)(0.050 m)(0.080 m) 0.025 A m IA = = = E VALUATE : The torque is a maximum when the field is in the plane of the loop and 90 = . 27.43. I DENTIFY : The period is 2 / T r v = , the current is / Q t and the magnetic moment is IA = S ET U P : The electron has charge e . The area enclosed by the orbit is 2 r . E XECUTE : (a) 16 2 1.5 10 s T r v  = = (b) Charge e passes a point on the orbit once during each period, so 1.1mA I Q t e t = = = . (c) 2 24 2 9.3 10 A m IA I r  = = = E VALUATE : Since the electron has negative charge, the direction of the current is opposite to the direction of motion of the electron. 27.44. I DENTIFY : sin IAB = , where is the angle between B r and the normal to the loop. S ET U P : The coil as viewed along the axis of rotation is shown in Figure 27.44a for its original position and in Figure 27.44b after it has rotated 30.0 ....
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 Fall '10
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