ε
==
=
d
dt
A
dB
dt
Ld
B
dt
B
Φ
loop
2
4
π
where the rate of change of the field is
dB
/
dt
= 0.0100 T/s. Consequently, we obtain
()
2
22 2
22
3
3
2
3
2
28
6
(
/ 4 ) (
/
)
(1.00 10 m) (0.500 m)
0.0100 T/s
/(
/ 4)
64
64 (1.69 10
m)
3.68 10 W .
B
d
t
d
L
d
B
P
RL
d
d
t
επ
ρπ
π
ρ
−
−
−
×
⎛⎞
=
=
⎜⎟
×Ω
⋅
⎝⎠
=×
29. Thermal energy is generated at the rate
P
=
2
/
R
(see Eq. 2723). Using Eq. 2716, the
resistance is given by
R =
L
/
A
, where the resistivity is 1.69
×
10
–8
Ω
·m (by Table 271)
and
A =
π
d
2
/4 is the crosssectional area of the wire (
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Resistance, Energy, Thermal Energy

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