Physics 212
Classical and Modern Physics
Spring 2008
Solutions for HandIn Set #3
Problem A13 — Resistance and Dimmer
Switches
(a)
I find a length of 22 cm, give or take a little.
(b)
From the assignment sheet we get the nichrome re
sistivity
ρ
= 1
.
0
×
10

6
Ω
·
m. Therefore the resistance
is
R
=
ρL
A
=
(1
.
0
×
10

6
Ω
·
m)(0
.
22 m)
π
(1
.
25
×
10

4
m)
2
=
4
.
5 Ω
(c)
Each battery provides 1
.
5 V, so in series they pro
vide a potential difference of 3 V. Using Ohm’s Law:
I
=
V
R
=
3 V
4
.
5 Ω
=
0
.
67 A
Problem A17 — Magnetic force on a moving
particle
The magnetic force is
F
=
qv
×
B
. The magnitude of
this force is
F
=

q

vB
sin
θ
, where
θ
is the angle be
tween
v
and
B
if I line them up tail to tail.
In this
problem
θ
= 60
◦
, so
F
= (5
×
10

6
C)(1
.
3
×
10
5
m)(0
.
8 T) sin(60
◦
)
=
0
.
450 N
As for direction,
v
×
B
points into the page, by the cross
product RHR. But since this is a negative charge, the
direction is flipped:
the force points out of the page.
Wolfson Chapter 2432
Power due to current flow is given by
P
=
V I
, always.
Since we’re given a resistance, we want to use Ohm’s
Law:
V
=
IR
. Solve for
I
=
V/R
and plug this into the
power expression to get
P
=
V
(
V/R
) =
V
2
/R
. This is
exactly what we need, since we’re given
P
and
R
, and
we want to find
V
:
V
=
√
PR
=
(1
.
5
×
10
3
W)(35 Ω) =
230 V
While the units in electricity and magnetism are messy,
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 Spring '08
 Ladd
 Resistance, Magnetic Field, ohm, RHR, Wolfson Chapter

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