816_PartUniversity Physics Solution

816_PartUniversity Physics Solution - Magnetic Field and...

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Magnetic Field and Magnetic Forces 27-23 sin sin60.0 B BI A B τμ φ == ° () gives sin60.0 0.0400 m sin30.0 Bm g IAB mg ττ = ° ( ) 3 0.15 g/cm 2 8.00 cm 6.00 cm 4.2 g 4.2 10 kg m =+ = = × ( ) 32 0.800 m 0.0600 m 4.80 10 m A × ( ) 0.0400 m sin30.0 sin60.0 mg B IA ° = ° ( )( ) 4.2 10 kg 9.80 m/s 0.0400 m sin30.0 0.024 T (8.2 A)(4.80 10 m )sin60.0 B ×° EVALUATE: As the loop swings up the torque due to B ! decreases to zero and the torque due to mg increases from zero, so there must be an orientation of the loop where the net torque is zero. 27.76. IDENTIFY: The torque exerted by the magnetic field is . = × B ! ! ! τ μ The torque required to hold the loop in place is . ! SET UP: . IA = ! is normal to the plane of the loop, with a direction given by the right-hand rule that is illustrated in Figure 27.32 in the textbook. sin , IAB = where is the angle between the normal to the loop and the direction of . B ! EXECUTE: (a) sin 60 (15.0 A)(0.060 m)(0.080 m)(0.48 T)sin60 0.030 N m τ IAB = ° = , in the & j direction. To keep the loop in place, you must provide a torque in the & + j direction. (b) sin 30 (15.0 A)(0.60 m)(0.080 m)(0.48 T)sin30 0.017 N m, τ IAB = ° = in the & + j direction. You must provide a torque in the & j direction to keep the loop in place. EVALUATE: (c) If the loop was pivoted through its center, then there would be a torque on both sides of the loop parallel to the rotation axis. However, the lever arm is only half as large, so the total torque in each case is identical to the values found in parts (a) and (b). 27.77. IDENTIFY: Use Eq.(27.20) to calculate the force and then the torque on each small section of the rod and integrate to find the total magnetic torque. At equilibrium the torques from the spring force and from the magnetic force cancel. The spring force depends on the amount x the spring is stretched and then 2 1 2 Uk x = gives the energy stored in the spring. (a) SET UP: Divide the rod into infinitesimal sections of length dr , as shown in Figure 27.77. Figure 27.77 EXECUTE: The magnetic force on this section is I dF IBdr = and is perpendicular to the rod. The torque d due to the force on this section is . I dr d FI B r d r The total torque is 2 1 2 0 0.0442 N m, clockwise. l dI B r d rI l B = ∫∫ (b) SET UP: I F produces a clockwise torque so the spring force must produce a counterclockwise torque. The spring force must be to the left; the spring is stretched. EXECUTE: Find x , the amount the spring is stretched: 0, = axis at hinge, counterclockwise torques positive 2 1 2 s i n 5 3 0 kx l Il B °− = ( ) ( ) 6.50 A 0.200 m 0.340 T 0.05765 m 2 sin53.0 2 4.80 N/m sin53.0 IlB x k = °° 23 1 2 7.98 10 J x × EVALUATE: The magnetic torque calculated in part (a) is the same torque calculated from a force diagram in which the total magnetic force I l B = acts at the center of the rod. We didn’t include a gravity torque since the problem said the rod had negligible mass.
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27-24 Chapter 27 27.78. IDENTIFY: Apply I × F=l B ! ! to calculate the force on each side of the loop.
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816_PartUniversity Physics Solution - Magnetic Field and...

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