P31_025.PDF

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Chapter 31 - 31.25
2 /R (see Eq. 27-23). Using Eq. 27-16, the resistance is
given by R = ρL/A , where the resistivity is 1 . 69
E
m (by Table 27-1) and A = πd 2 / 4isthe
cross-sectional area of the wire ( d =0 . 00100 m is the wire thickness). The area enclosed by the loop is
A loop = πr loop = π L
×
10 − 8 Ω
·
2
2 π
since the length of the wire ( L =0 . 500 m) is the circumference of the loop. This enclosed area is used in
Faraday’s law (where we ignore minus signs in the interest of finding the magnitudes of the quantities):
E
= d Φ B
dt
= A loop dB
dt = L 2
4 π
dB
dt
where the rate of change of the field is dB/dt =0 . 0100 T/s. Consequently, we obtain
4 ρL/πd 2 = d 2 L 3
dB
dt 2
dB
dt
2
P =
=3 . 68
×
10 − 6 W .
64 πρ
25. Thermal energy is generated at the rate P =
L 2
4 π
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