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856_PartUniversity Physics Solution

856_PartUniversity Physics Solution - 29-6 Chapter 29 SET...

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29-6 Chapter 29 S ET U P : The loop when it is completely inside the field region is sketched in Figure 29.26b. E XECUTE : For /2 /2 L x L < < the loop is completely inside the field region and 2 . B BL Φ = Figure 29.26b But 0 so 0 and 0. B d I dt Φ = = = E There is no force I F = l B ! ! ! × from the magnetic field and the external force F necessary to maintain constant velocity is zero. S ET U P : The loop as it enters the magnetic field region is sketched in Figure 29.26c. E XECUTE : For 3 /2 /2 L x L < < − the loop is entering the field region. Let x be the length of the loop that is within the field. Figure 29.26c Then and . B B d BLx Blv dt Φ Φ = = The magnitude of the induced emf is B d BLv dt Φ = = E and the induced current is . BLv I R R = = E Direction of I : Let A ! be directed into the plane of the figure. Then B Φ is positive. The flux is positive and increasing in magnitude, so B d dt Φ is positive. Then by Faraday°s law E is negative, and with our choice for direction of A ! a negative E is counterclockwise. The current induced in the loop is counterclockwise. S ET U P : The induced current and magnetic force on the loop are shown in Figure 29.26d, for the situation where the loop is entering the field. E XECUTE : I I = F l B ! ! ! × gives that the force I F ! exerted on the loop by the magnetic field is to the left and has magnitude 2 2 . I BLv B L v F ILB LB R R = = = Figure 29.26d The external force F ! needed to move the loop at constant speed is equal in magnitude and opposite in direction to I F ! so is to the right and has this same magnitude. S ET U P : The loop as it leaves the magnetic field region is sketched in Figure 29.26e. E XECUTE : For /2 3 /2 L x L < < the loop is leaving the field region. Let x be the length of the loop that is outside the field. Figure 29.26e Then ( ) and . B B d BL L x BLv dt Φ Φ = = The magnitude of the induced emf is B d BLv dt Φ = = E and the induced current is . BLv I R R = = E Direction of I : Again let A ! be directed into the plane of the figure. Then B Φ is positive and decreasing in magnitude, so B d dt Φ is negative. Then by Faraday°s law E is positive, and with our choice for direction of A ! a positive E is clockwise. The current induced in the loop is clockwise.
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Electromagnetic Induction 29-7 S ET U P : The induced current and magnetic force on the loop are shown in Figure 29.26f, for the situation where the loop is leaving the field. E XECUTE : I I = F l B ! ! ! × gives that the force I F ! exerted on the loop by the magnetic field is to the left and has magnitude 2 2 . I BLv B L v F ILB LB R R = = = Figure 29.26f The external force F ! needed to move the loop at constant speed is equal in magnitude and opposite in direction to I F !
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