# ass20 - Solutions A.42 I DENTIFY sin IAB τ φ = The...

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Unformatted text preview: Solutions, Assignment 20 27.42. I DENTIFY : sin IAB τ φ = . The magnetic moment of the loop is IA μ = . 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 φ = ° . E XECUTE : (a) 3 (6.2 A)(0.050 m)(0.080 m)(0.19 T) 4.7 10 N m IAB τ- = = = × ⋅ (b) 2 (6.2 A)(0.050 m)(0.080 m) 0.025 A m IA μ = = = ⋅ E VALUATE : The torque is a maximum when the field is in the plane of the loop and 90 φ = ° . 27.43. I DENTIFY : The period is 2 / T r v π = , the current is / Q t and the magnetic moment is IA μ = S ET U P : The electron has charge e- . The area enclosed by the orbit is 2 r π . E XECUTE : (a) 16 2 1.5 10 s T r v π- = = × (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 μ π- = = = × ⋅ 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 τ φ = , where φ 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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ass20 - Solutions A.42 I DENTIFY sin IAB τ φ = The...

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