PHYS1002 TUTORIAL 5 SOLUTIONS
Semester 2, 2007
1.
MAGNETIC MATERIALS
We have
B
τ
μ
=
×
arrowrightnosp
arrowrightnosp arrowrightnosp
where
μ
arrowrightnosp
is the magnetic moment of the magnet,
B
arrowrightnosp
is the magnetic field
of the solenoid, and
τ
arrowrightnosp
is the torque experienced by the magnet. Here the magnetic moment
will be along the length of the magnet and hence perpendicular to
B
arrowrightnosp
. So
τ
μ
=
B
and therefore
/
B
μ
τ
=
.
The magnetic field of the solenoid is given by
B
n I
o
= μ
where
n
is the number of turns per meter and
I
is the current through the solenoid.
Here
B
T
=
×
×
×
=
×
−
−
4
10
1000
10
1 26
10
7
2
π
.
Hence
μ =
×
×
=
−
−
18
10
1 26
10
1 43
2
2
2
.
/ .
.
A m
The magnetisation is the magnetic moment per unit volume,
M
d
dV
A m
=
=
×
×
×
=
×
−
μ
π
1 43
0 1
5
10
7 3
10
3
2
5
.
.
.
/
e j
Since
M
di
d
=
ℓ
where
i
is the Amperean current and
ℓ
is the length of the magnet, we have
i
A
=
×
×
=
×
7 3
10
0 1
7 3
10
5
4
.
.
.
.
2.
FARADAY’S LAW I
Let
ˆ
n
be a unit vector perpendicular to the plane of the loop. At time t = 0, we choose
ˆ
n
to
point in the
positive
z direction. Let
θ
be the angle between
B
arrowrightnosp
and ˆ
n
. Since the loop is set to
rotate at a
constant
angular frequency
ω
and since
0
θ
=
at
t = 0, it follows that
t
θ
ω
=
for
0
t
≥
. The magnetic flux through the loop is then
2
ˆ
cos
cos(
)
S
B ndA
BA
B
r
t
φ
θ
π
ω
=
⋅
=
=
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 Three '11
 TarasPlank
 Magnetism, Magnetic Field, magnetic materials

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