Solution for Homework 14 Introduction to Magnetic Fields
Solution to Homework Problem 14.1(Magnetic Field of Pith Ball)
Problem:
A pith ball is charged with an electrophorus to
1nC
and shot along the
x
axis at
100
m
s
in the
+
x
direction. What is the magnitude of the magnetic field at
(0
,
1cm
,
0)
when it reaches the origin?
Select One of the Following:
(aAnswer)
1
×
10
−
10
Tˆ
z
(b)
2
×
10
−
10
Tˆy
(c)
4
×
10
−
7
Tˆ
z
(d)
3
×
10
−
11
Tˆ
z
(e)
2
×
10
−
10
Tˆx
Solution
The magnetic field of a moving charge is
vector
B
=
μ
0
4
π
qvectorv
×
ˆ
r
10
r
2
10
where the vector
vector
r
10
points from the location of the charge to the location where the field is computed,
q
= 1nC
is the charge,
vectorv
= 100
m
s
ˆx
is the velocity. When the charge reaches the origin,
vector
r
10
= (0
,
1cm
,
0)
,
r
10
= 1cm
, and
ˆ
r
10
= ˆ
y
. Compute the cross product,
vectorv
×
ˆ
r
10
= 100
m
s
(ˆx
×
ˆy) = 100
m
s
ˆ
z
, where we use the right hand rule to solve
the cross product. Substitute into the complete expression for
vector
B
giving
vector
B
=
μ
0
4
π
q
r
2
10
(
vectorv
×
ˆ
r
10
) =
parenleftbigg
4
π
×
10
−
7 Tm
A
4
π
parenrightbiggparenleftbigg
1
×
10
−
9
C
(0
.
01m)
2
parenrightbiggparenleftbigg
100
m
s
ˆ
z
parenrightbigg
vector
B
= 1
×
10
−
10
Tˆ
z
2 point(s) : Magnitude
1 point(s) : Direction.
Total Points for Problem: 3 Points
Solution to Homework Problem 14.2(Right Handed Coordinate Systems)
Problem:
If the
y
axis points to the right of the page and the
z
axis points to the top of the page, what
direction does the
x
axis point?
Select One of the Following:
(a) To the left of the page.
(b) To the right of the page.
(c) To the top of the page.
(d) To the bottom of the page.
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 Spring '10
 Stewat
 Charge, Work, Magnetic Field

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