806_PartUniversity Physics Solution

806_PartUniversity Physics Solution - Magnetic Field and...

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Magnetic Field and Magnetic Forces 27-13 27.44. IDENTIFY: sin IAB τ φ = , where is the angle between B ! and the normal to the loop. SET UP: 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 ° . EXECUTE: (a) The forces on each side of the coil are shown in Figure 27.44a. 12 0 + = FF ! ! and 34 0 += !! . The net force on the coil is zero. 0 = ° and sin 0 = , so 0 = . The forces on the coil produce no torque. (b) The net force is still zero. 30.0 = ° and the net torque is (1)(1.40 A)(0.220 m)(0.350 m)(1.50 T)sin30.0 0.0808 N m == ° . The net torque is clockwise in Figure 27.44b and is directed so as to increase the angle . EVALUATE: For any current loop in a uniform magnetic field the net force on the loop is zero. The torque on the loop depends on the orientation of the plane of the loop relative to the magnetic field direction. Figure 27.44 27.45. IDENTIFY: The magnetic field exerts a torque on the current-carrying coil, which causes it to turn. We can use the rotational form of Newton&s second law to find the angular acceleration of the coil. SET UP: The magnetic torque is given by = × B ! ! ! τμ , and the rotational form of Newton&s second law is I α = . The magnetic field is parallel to the plane of the loop. EXECUTE: (a) The coil rotates about axis A 2 because the only torque is along top and bottom sides of the coil. (b) To find the moment of inertia of the coil, treat the two 1.00-m segments as point-masses (since all the points in them are 0.250 m from the rotation axis) and the two 0.500-m segments as thin uniform bars rotated about their centers. Since the coil is uniform, the mass of each segment is proportional to its fraction of the total perimeter of the coil. Each 1.00-m segment is 1/3 of the total perimeter, so its mass is (1/3)(210 g) = 70 g = 0.070 kg. The mass of each 0.500-m segment is half this amount, or 0.035 kg. The result is 22 2 1 12 2(0.070 kg)(0.250 m) 2 (0.035 kg)(0.500 m) 0.0102 kg m I =+ = The torque is = sin90 (2.00 A)(0.500 m)(1.00 m)(3.00 T) = 3.00 N m IAB ° = B ! Using the above values, the rotational form of Newton&s second law gives 2 290 rad/s I EVALUATE: This angular acceleration will not continue because the torque changes as the coil turns. 27.46. IDENTIFY: B ! and cos UB μ =− , where NIB = . sin B μφ = . SET UP: is the angle between B ! and the normal to the plane of the loop. EXECUTE: (a) ±± ± 90 . sin(90 ) , direction . cos 0. τ NIAB NIAB U B φμ =° = ° = × − = = kj =i (b) 0. sin(0) 0, no direction. cos . τ NIAB U B NIAB = = = (c) ±±± 90 . sin(90 ) , direction . cos 0. τ NIAB NIAB U B ° = −× = = = i (d) 180 : sin(180 ) 0, no direction, cos(180 ) . τ NIAB U B NIAB = ° = = ° = EVALUATE: When is maximum, 0 U = . When U is maximum, 0 = . 27.47. IDENTIFY and SET UP: The potential energy is given by Eq.(27.27): . U = B ! ! The scalar product depends on the angle between and . B ! ! EXECUTE: For and parallel, 0 and cos . B B ⋅ = = BB μμ For and antiparallel, B ! ! 180 and cos . B B = = !
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806_PartUniversity Physics Solution - Magnetic Field and...

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