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871_PartUniversity Physics Solution

# 871_PartUniversity Physics Solution - Electromagnetic...

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Electromagnetic Induction 29-21 29.65. (a) I DENTIFY : Use Faraday°s law to calculate the induced emf, Ohm°s law to calculate I , and Eq.(27.19) to calculate the force on the rod due to the induced current. S ET U P : The force on the wire is shown in Figure 29.65. E XECUTE : When the wire has speed v the induced emf is Bva = E and the induced current is / Bva I R R = = E Figure 29.65 The induced current flows upward in the wire as shown, so the force I F = l B ! ! ! × exerted by the magnetic field on the induced current is to the left. F ! opposes the motion of the wire, as it must by Lenz°s law. The magnitude of the force is 2 2 / . F IaB B a v R = = (b) Apply m F a ! ! = to the wire. Take + x to be toward the right and let the origin be at the location of the wire at t = 0, so 0 0. x = says x x x F ma F ma = = 2 2 x F B a v a m mR = − = − Use this expression to solve for v ( t ): 2 2 2 2 and x dv B a v dv B a a dt dt mR v mR = = − = − 0 2 2 0 v t v dv B a dt v mR = − 2 2 0 ln( ) ln( ) B a t v v mR = − 2 2 2 2 / 0 0 ln and B a t mR v B a t v v e v mR = − = Note: At 0 0, and 0 when t v v v t = = → ∞ Now solve for x ( t ): 2 2 2 2 / / 0 0 so B a t mR B a t mR dx v v e dx v e dt dt = = = 2 2 / 0 0 0 x t B a t mR dx v e dt = ( ) 2 2 2 2 / / 0 0 2 2 2 2 0 1 t B a t mR B a t mR mR mRv x v e e B a B a = = Comes to rest implies v = 0. This happens when . t → ∞ 0 2 2 gives . mRv t x B a → ∞ = Thus this is the distance the wire travels before coming to rest. E VALUATE : The motion of the slide wire causes an induced emf and current. The magnetic force on the induced current opposes the motion of the wire and eventually brings it to rest. The force and acceleration depend on v and are constant. If the acceleration were constant, not changing from its initial value of 2 2 0 / , x a B a v mR = − then the stopping distance would be 2 2 2 0 0 /2 /2 . x x v a mRv B a = − = The actual stopping distance is twice this. 29.66. I DENTIFY : Since the bar is straight and the magnetic field is uniform, integrating d d ε = × v B l ! ! ! along the length of the bar gives ( ) = × v B L ! ! ! E S ET U P : ± (4.20 m/s) v = i ! . ± ± (0.250 m)(cos36.9 sin36.9 ). + L = i j ! ° ° E XECUTE : (a) ( ) ± ± ± ± ( ) (4.20 m/s) ((0.120 T) 0.220 T (0.0900 T) ) . = × = × v B L i i j k L ! ! ! ! E ( ) ( ) ( ) ( ) ± ± ± ± 0.378 V/m 0.924 V/m (0.250 m)(cos 36.9 sin36.9 ) . = ° + ° ° j k i j E (0.378 V/m)(0.250 m)sin36.9 0.0567 V. = ° = E (b) The higher potential end is the end to which positive charges in the rod are pushed by the magnetic force. × v B ! ! has a positive y -component, so the end of the rod marked + in Figure 29.66 is at higher potential.

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29-22 Chapter 29 E VALUATE : Since × v B ! ! has nonzero ± j and ± k components, and L ! has nonzero ± i and ± j components, only the ± k component of B ! contributes to . E In fact, (4.20 m/s)(0.0900 T)(0.250 m)sin36.9 0.0567 V.
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871_PartUniversity Physics Solution - Electromagnetic...

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