Induced Voltages and Inductance
179
Problem Solutions
20.1
The magnetic flux through the area enclosed by the loop is
( )
()
( )
2
22
cos
cos0
0.30 T
0.25 m
5.9
10
T m
B
BA
B
r
θπ
π
−
Φ=
=
°=
=
×
⋅
2
20.2
The magnetic flux through the loop is given by
cos
B
BA
θ
Φ
=
where
B
is the magnitude
of the magnetic field,
A
is the area enclosed by the loop, and
is the angle the magnetic
field makes with the normal to the plane of the loop. Thus,
52
7
2
10
cos
5.00
10
T
20.0 cm
1.00
10
T m
cos
B
BA
2
2
m
cos
1 cm
θθ
−−
=
×
×
⋅
=
(a) When
B
is perpendicular to the plane of the loop,
JG
0
=
°
and
72
10
T m
B
−
×
⋅
1.00
(b) If
8
30.0 , then
1.00
10
T m
cos30.0
8.66
10
T m
B
Φ
= ×
⋅
°
⋅
2
=°
(c) If
( )
90.0 , then
1.00
10
T m
cos90.0
0
B
−
Φ
⋅
°
=
20.3
The magnetic flux through the loop is given by
cos
B
BA
Φ
=
where
B
is the magnitude
of the magnetic field,
A
is the area enclosed by the loop, and
is the angle the magnetic
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.
 Fall '10
 STAFF
 Physics, Inductance

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