α
cos
2
α
006
(part 2 of 3) 10 points
What is the power lost to dissipation in the
circuit?
1.
P
=
1
R
(
B ‘ v
sin
α
)
2
2.
P
=
B
2
‘
2
v
2
R
cos
α
sin
2
α
3.
P
=
1
R
(
B ‘ v
cos
α
)
2
correct
4.
P
=
B ‘ v
R
sin
α
5.
P
=
B
2
‘
2
R
cos
2
α
6.
P
=
B ‘ v
R
sin
2
α
Explanation:
For current
I
and resistance
R
, the power
dissipation is
P
=
I
2
R .
And from (4) of Part 1
P
=
I
2
R
=
1
R
B
⊥
‘ v
¶
2
R
=
B
2
⊥
‘
2
v
2
R
=
B
2
‘
2
v
2
R
cos
2
α .
Husain, Zeena – Homework 10 – Due: Apr 5 2004, 4:00 am – Inst: Sonia Paban
4
007
(part 3 of 3) 10 points
(denote the original steadystate speed as
v
1
)
Reverse the direction of
B
.
The new steadystate speed
v
2
is
009
(part 2 of 3) 10 points
The direction of the induced magnetic field at
the center of the circulating eddy current is
Explanation:
O
008
(part 1 of 3) 10 points
A pendulum consists of a supporting rod
and a metal plate (see figure).
The rod is
pivoted at
O
. The metal plate swings through
a region of magnetic field. Consider the case
where the pendulum is entering the magnetic
field region from the left.
O
enter
010
(part 3 of 3) 10 points
The direction of the force which the magnetic
field exerts is
1.
along the direction of swing.
2.
opposite to the direction of swing.
cor
rect
3.
along the rod toward the pivot point.
4.
out of the plane.
5.
into the plane.
F
B
i
Explanation:
Husain, Zeena – Homework 10 – Due: Apr 5 2004, 4:00 am – Inst: Sonia Paban
5
Because the magnetic field only exerts a
force on the current segment already in the
magnetic field region, the net magnetic force
is opposite to the direction of swing, see the
figure in the explanation of the previous Part.
011
(part 1 of 1) 10 points
A coiled telephone cord has 85 turns, a cross
sectional diameter of 4
.
21 cm, and an un
stretched length of 32 cm.
Determine an
approximate value for the selfinductance of
the unstretched cord.
012
(part 1 of 1) 10 points
Given:
The
resistivity
of
copper
is
1
.
7
×
10

8
Ω m
.