P27_041.PDF

41. (a) The charge q that ﬂows past any cross section of the beam in time ∆ t is given by q = i ∆ t ,andthe

number of electrons is N = q/e =( i/e )∆ t . This is the number of electrons that are accelerated.

Thus

10 − 6 s)

N = (0 . 50A)(0 . 10

=3 . 1

10 11

1 . 60

10 − 19 C

(b) Over a long time t the total charge is Q = nqt ,where n is the number of pulses per unit time and

q is the charge in one pulse. The average current is given by i avg = Q/t = nq .Now q = i ∆ t =

(0 . 50A)(0 . 10

10 − 6 s) = 5 . 0

10 − 8 C, so

i avg =(500 / s)(5 . 0

10 − 8 C) = 2 . 5

10 − 5 A .

Since K = 50MeV, the accelerating potential is V =50kV=5 . 0

10 7 V. During a pulse the power

output is

P = iV =(0 . 50A)(5 . 0

10 7 V) = 2 . 5

10 7 W .

This is the peak power. The average power is

P avg = i avg V =(2 . 5

10 − 5 A)(5 . 0

10 7 V) = 1 . 3

10 3 W .