P30_028.PDF

28. (a) Consider a segment of the projectile between y and y + dy . We use Eq. 30-14 to ﬁnd the magnetic

force on the segment, and Eq. 30-9 for the magnetic ﬁeld of each semi-inﬁnite wire (the top rail

referred to as wire 1 and the bottom as wire 2). The current in rail 1 is in the + i direction, and

the current in the rail 2 is in the

−

i direction. The ﬁeld (in the region between the wires) set up

k direction) and that set up by wire 2 is also into the paper. The

force element (a function of y ) acting on the segment of the projectile (in which the current ﬂows

in the

by wire 1 is into the paper (the

−

j direction) is given below. The coordinate origin is at the bottom of the projectile.

d F = d F 1 + d F 2

= idy (

−

B 1 + dy (

−

B 2

= i [ B 1 + B 2 ]i dy

= i µ 0 i

4 π (2 R + w

−

y ) + µ 0 i

4 πy

i dy .

Thus, the force on the projectile is

F =

d F = i 2 µ 0

4 π

R + w

2 R + w − y + 1

dy ˆ i= µ 0 i 2

2 π

ln 1+ w

i .

(b) Using the work-energy theorem, we have ∆ K = 2 mv f = W ext =

ds = FL . Thus, the ﬁnal

speed of the projectile is

2 W ext

1 / 2

= 2

µ 0 i 2

2 π

ln 1+ w

L 1 / 2

v f =

2(4 π

1 / 2

10 − 7 T

m / A)(450

10 3 A) 2 ln(1 + 1 . 2cm / 6 . 7 cm)(4 . 0m)

2 π (10

10 − 3 kg)

=2 . 3

10 3 m / s .

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