(b) If
Φ
is the magnetic flux through the secondary coil, then the magnitude of the emf
induced in that coil is
ε
=
N
(
d
Φ
/
dt
) and the current in the secondary is
i
s
=
/
R
, where
R
is
the resistance of the coil. Thus,
i
N
R
d
dt
s
=
F
H
G
I
K
J
Φ
.
The charge that passes through the secondary when the primary current is turned on is
0
.
s
Nd
N
N
q
i dt
dt
d
R
dt
R
R
Φ
Φ
Φ
==
=
Φ
=
∫∫
∫
The magnetic field through the secondary coil has magnitude
B = B
0
+
B
M
= 801
B
0
,
where
B
M
is the field of the magnetic dipoles in the magnetic material. The total field is
perpendicular to the plane of the secondary coil, so the magnetic flux is
Φ
=
AB
, where
A
is the area of the Rowland ring (the field is inside the ring, not in the region between the
ring and coil). If
r
is the radius of the ring’s cross section, then
A =
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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, Current

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