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q = 2e = 2(1.60 × 10–19 C) = 3.20 × 10–19 C,
and the drift speed is 1.0 × 105 m/s. Thus, c hc hc h J = 2 × 1014 / m 3.2 × 10−19 C 10 × 105 m / s = 6.4 A / m2 .
.
(b) Since the particles are positively charged the current density is in the same direction
as their motion, to the north.
(c) The current cannot be calculated unless the crosssectional area of the beam is known.
Then i = JA can be used.
6. (a) Circular area depends, of course, on r2, so the horizontal axis of the graph in Fig.
2623(b) is effectively the same as the area (enclosed at variable radius values), except
for a factor of π. The fact that the current increases linearly in the graph means that i/A =
J = constant. Thus, the answer is “yes, the current density is uniform.”
(b) We find i/(πr2) = (0.005 A)/(π × 4 × 10−6 m2) = 398 ≈ 4.0 × 102 A/m2.
7. The crosssectional area of wire is given by A = πr2, where r is its radius (half its
thickness). The magnitude of the current density vector is
J = i / A = i / π r2,
so 1041 i
0.50 A
=
= 1.9 ×10−4 m.
4
2
πJ
π ( 440 ×10 A/m ) r= The diameter of the wire is therefore d = 2r = 2(1.9 × 10–4 m) = 3.8 × 10–4 m.
8. (a) The magnitude of the current density vector is
4 (1.2 ×10−10 A )
i
i
J= =
=
= 2.4 ×10−5 A/m 2 .
2
2
−3
A π d / 4 π ( 2.5 ×10 m )
(b) The drift speed of the currentcarrying electrons is
J
2.4 × 10−5 A / m2
vd =
=
= 18 × 10−15 m / s.
.
28
3
−19
ne 8.47 × 10 / m 160 × 10 C
. c hc h 9. We note that the radial width Δr = 10 μm is small enough (compared to r = 1.20 mm)
that we can make the approximation ∫ Br 2π rdr ≈ Br 2π r Δr
Thus, the enclosed current is 2πBr2Δr = 18.1 μA. Performing the integral gives the same
answer.
10. Assuming J is directed along the wire (with no radial flow) we integrate, starting
with Eq. 264,
R
1
(kr 2 )2π rdr = kπ ( R 4 − 0.656 R 4 )
i = ∫  J  dA = ∫
9 R /10
2
where k = 3.0 × 108 and SI units are understood. Therefore, if R = 0.00200 m, w...
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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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