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the possibility of an overall minus sign since we are asked to find the absolute value of
the flux.
/2
00

()
l
n
.
22
/
2
rb
B
ii
a
adr
rr
b
µµ
+
−
+
⎛⎞
Φ=
=
⎜⎟
ππ
−
⎝⎠
∫
When
1.5
=
, we have
8
(4
T m A)(4.7A)(0.022m)


ln(2.0)
1.4 10 Wb.
2
B
π
−
×⋅
= ×
−7
π1
0
(b) Implementing Faraday’s law involves taking a derivative of the flux in part (a), and
recognizing that
/
dr dt
v
=
. The magnitude of the induced emf divided by the loop
resistance then gives the induced current:
loop
3
42
5
ln
2/
2
2
[
(
/
2
)
]
(4
T m A)(4.7A)(0.022m)(0.0080m)(3.2 10 m/s)
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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

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