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Summary 849 (B) What is the average current per pulse delivered by the accelerator? Solution Average current is given by Equation 27.1, av " ! Q /!t. Because the time interval between pulses is 4.00 ms, and because we know the charge per pulse from This represents only 0.005% of the peak current, which is (C) What is the peak power delivered by the electron beam? Solution By definition, power is energy delivered per unit 10.0 MW " 1.00 # 10 7 W " # " 1.60 # 10 $ 13 J 1 MeV # " (3.13 # 10 11 electrons/pulse)(40.0 MeV/electron) 2.00 # 10 $ 7 s/pulse (1)
" peak " pulse energy pulse duration 12.5 /A I av " Q pulse ∆t " 5.00 # 10 $ 8 C 4.00 # 10 $ 3 s " 3.13 # 10 11 electrons/pulse " Electrons per pulse " 5.00 # 10 $ 8 C/pulse 1.60 # 10 $ 19 C/electron We could also compute this power directly. We assume that What If? What if the requested quantity in part (C) were the average power rather than the peak power? Answer Instead of Equation (1), we would use the time Instead of Equation (2), we would use the average current Notice that these two calculations agree with each other and " 500 W " av " I av
∆V " (12.5 # 10 $ 6 A)(40.0 # 10 6 V
) " 500 W # " 1.60 # 10 $ 13 J 1 MeV # " (3.13 # 10 11 electrons/pulse)(40.0 MeV/electron) 4.00 # 10 $ 3 s/pulse " av " pulse energy time interval between pulses 10.0 MW " " (250 # 10 $ 3 A)(40.0 # 10 6 V
) (2)
" peak " I peak ∆V The electric current I in a conductor is defined as (27.2) where dQ is the charge that passes through a cross section of the conductor in a time ampere (A), where 1 A " 1 C/s. The average current in a conductor is related to the motion of the charge carriers through the relationship (27.4) where n is the density of charge carriers, q is the charge on each carrier, v d is the drift speed, and A is the cross-sectional area of the conductor. The magnitude of the current density J in a conductor is the current per unit area: (27.5) J ! I A " nqv d I av " nqv d A I ! dQ dt S U M M A R Y Take a practice test for this chapter by clicking on |