This preview shows pages 1–3. Sign up to view the full content.
PS08 Solutions1
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2008
Problem Set 8 Solutions
Problem 1: Charges in a Uniform Field
An electron with velocity
ˆˆ
xy
vv
+
i
j
moves through a uniform magnetic field
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 crossproduct.
Although it wasn’t introduced in
class, the way to do this is with a determinant:
()
ˆ
ˆ
0
0
x
y
y
x
qq
v
v
q
v
B
v
B
BB
=×
=
=
−
ij
k
Fv
B
k
GG
G
So the magnitude of the force is
( )
yx
Fe
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
= 2000 A,
= 73.0º 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
IF
I
L
B
⇒
=
FL
B
G
For the numbers given,
( ) ( )( )(
)()
sin
2000 A 100 m 62.1
sin 73º
12 N
FI
L
B
T
θµ
==
=
,
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 3, 4, and
5 cm. The loop is in a uniform magnetic field
B
whose direction is parallel to the current
in the 5 cm side of the loop. Find the magnitude of
(a)
the magnetic dipole moment of the
loop in amperessquare meters and
(b)
the torque on the loop.
The loop has an area of
2
1
2
base height
6 cm
A
=⋅
=
so its dipole moment is
(a)
2
6
c
m
I
µ
(b)
The torque is
2
6
cm (in the plane of the loop, perpendicular to the hypotenuse)
IB
=×=
τ
μB
G
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentPS08 Solutions2
Problem 4:
A charge of mass
m
and charge
q > 0
is at the origin at
t = 0
and moving upward with
velocity
ˆ
V
=
V
j
G
.
Its subsequent trajectory is shown in the sketch.
The magnitude of the
velocity
V
=
V
G
is always the same, although the direction of
V
G
changes in time.
(a)
For
y > 0
, this positive charge is moving in a constant magnetic field which is either
into the page or out of the page.
Which is it?
Into the Page
(b)
Derive an expression for the magnitude of the magnetic field for
y > 0
.
We just balance centripetal and magnetic force:
2
q
V
Bm
VR
Vq
R
=⇒
=
(c)
For
y < 0
, the charge is moving in a
different
constant magnetic field.
What is that
magnetic field (magnitude and direction)?
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 SCIOLLA

Click to edit the document details