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obtain i = 2.59 ×10−3 A .
11. (a) The current resulting from this nonuniform current density is
i=∫ cylinder J a dA = = 1.33 A.
(b) In this case, J0
R ∫ R 0 2
2
r ⋅ 2π rdr = π R 2 J 0 = π (3.40 ×10−3 m) 2 (5.50 ×104 A/m 2 )
.
3
3 1042 CHAPTER 26 R
1
1
⎛ r⎞
J b dA = ∫ J 0 ⎜1 − ⎟ 2π rdr = π R 2 J 0 = π (3.40 ×10−3 m) 2 (5.50 ×104 A/m 2 )
cylinder
0
3
3
⎝ R⎠
= 0.666 A. i=∫ (c) The result is different from that in part (a) because Jb is higher near the center of the
cylinder (where the area is smaller for the same radial interval) and lower outward,
resulting in a lower average current density over the cross section and consequently a
lower current than that in part (a). So, Ja has its maximum value near the surface of the
wire.
12. (a) Since 1 cm3 = 10–6 m3, the magnitude of the current density vector is
J = nev = FG 8.70 IJ c160 × 10 Chc470 × 10 m / sh = 6.54 × 10
H 10 m K .
−19 −6 3 3 −7 A / m2 . 2
(b) Although the total surface area of Earth is 4π RE (that of a sphere), the area to be used
in a computation of how many protons in an approximately unidirectional beam (the solar
wind) will be captured by Earth is its projected area. In other words, for the beam, the
2
encounter is with a “target” of circular area π RE . The rate of charge transport implied by
the influx of protons is
2
i = AJ = π RE J = π ( 6.37 ×106 m ) ( 6.54 ×10−7 A/m 2 ) = 8.34 ×107 A.
2 13. We use vd = J/ne = i/Ane. Thus,
−14
2
28
3
−19
L
L
LAne ( 0.85m ) ( 0.21× 10 m ) ( 8.47 × 10 / m ) (1.60 × 10 C )
t=
=
=
=
300A
vd i / Ane
i = 8.1× 102 s = 13min .
14. Since the potential difference V and current i are related by V = iR, where R is the
resistance of the electrician, the fatal voltage is V = (50 × 10–3 A)(2000 Ω) = 100 V.
15. The resistance of the coil is given by R = ρL/A, where L is the length of the wire, ρ is
the resistivity of copper, and A is the crosssectional area of the wire. Since each turn of
wire has length 2πr,...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.
 Fall '08
 schuller
 Magnetism, Work

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