This preview shows page 1. Sign up to view the full content.
10. (a) Application of Eq. 323 along the circle referred to in the second sentence of the
problem statement (and taking the derivative of the flux expression given in that sentence)
leads to
()
00
(2
)
0.60 V m/s
r
Br
R
πε
µ
=⋅
.
Using
r
= 0.0200 m (which, in any case, cancels out) and
R
= 0.0300 m, we obtain
12
2
2
7
17
(0.60 V m/s)
(8.85 10
C /N m )(4
10 T m/A)(0.60 V m/s)
2
2 (0.0300 m)
3.54 10
T .
B
R
εµ
ππ
−−
−
⋅
×⋅
π
×
⋅
⋅
==
=×
(b) For a value of
r
larger than
R
, we must note that the flux enclosed has already reached
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details