MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2007
Problem Set 8 Solutions
Problem 1: Charges in a Uniform Field
An electron with velocity
ˆ
ˆ
x
y
v
v
+
i
j
moves through a uniform magnetic field
ˆ
ˆ
x
y
B
B
+
i
j
(a) Find the magnitude of the force on the electron. (b) Repeat your calculation for a
proton having the same velocity.
This is just a question of calculating a cross-product.
Although it wasn’t introduced in
class, the way to do this is with a determinant:
(
)
ˆ
ˆ
ˆ
ˆ
0
0
x
y
x
y
y
x
x
y
q
q v
v
q v B
v B
B
B
=
×
=
=
−
i
j
k
F
v
B
G
G
G
k
So the magnitude of the force is
(
)
x
y
y
x
F
e v B
v B
=
−
for both an electron and proton, the
only difference is the direction is reversed (
q
=-
e
for an electron).
Problem 2: Power Line
A horizontal power line carries a current of
I
from south to north. Earth's magnetic field
(
B
Earth
= 62.1 μT) is directed toward the north and is inclined downward at an angle
θ
to
the horizontal. Find the magnitude of the magnetic force on a length
L
of the line due to
Earth's field.
Please also calculate a value numerically for
I
= 3000 A,
θ
= 78.0º and
L
=
100 m, since it’s useful to think about real world numbers.
Since the line is running south to north, it has an angle
θ
relative to the Earth’s field.
Thus:
(
)
sin
I
F
ILB
θ
=
×
⇒
=
F
L
B
G
G
G
For the numbers given,
(
)
(
)(
)(
)
(
)
sin
3000 A
100 m
62.1
sin
78º
18 N
F
ILB
T
θ
μ
=
=
=
,
which is a small force.
For comparison, if the cable has a mass of 300 kg (about what a
one cm radius copper wire of that length will weigh) it will feel a force of about 3000 N
due to gravity.
Of course most power lines are AC so this will also be an oscillating, not
unidirectional force.
Problem 3: Triangular Loop
A current loop, carrying a current
I
, is in the shape of a right triangle with sides 30, 40,
and 50 cm. The loop is in a uniform magnetic field
B
whose direction is parallel to the
current in the 50 cm side of the loop. Find the magnitude of
(a)
the magnetic dipole
moment of the loop in amperes-square meters and
(b)
the torque on the loop.
The loop has an area of
2
1
2
base
height
600 cm
A
=
⋅
=
so its dipole moment is
(a)
2
1200
cm
I
μ
=
⋅
(b)
The torque is
2
600
cm
(in the plane of the loop, perpendicular to the hypotenuse)
IB
=
×
=
τ
μ
B
G
G
G
PS08 Solutions-1

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