Q v b sin θ 1 60218 10 19 c3 77 10 6 m s 0 308 t sin

= q v B sin θ
= (1 . 60218 × 10
− 19
C)(3 . 77 × 10
6
m / s)
× (0 . 308 T) sin 19 . 8
◦
= 6 . 30183 × 10
− 14
N.
006(part2of2)10.0points
The mass of proton is 1 . 67262 × 10
− 27
kg
What is the proton’s acceleration?
Correct answer: 3 . 76763 × 10
13
m / s
2
.
Explanation:
Let : m = 1 . 67262 × 10
− 27
kg ,
F = m a .
a =
F
m
=
6 . 30183 × 10
− 14
N
1 . 67262 × 10
− 27
kg
= 3 . 76763 × 10
13
m / s
2
.
007 10.0points
A indicates the magnetic field B point-
ing up from the surface of the paper and
a indicates the magnetic field B pointing
down into the surface of the paper.
A copper bar has a constant velocity v in
the plane of the paper and is perpendicular to
a magnetic field B as shown in the figures.
How are charges distributed on the bar?
1.
v
B
B
+
+
-
-

gutierrez (bg25688) – Hw9 – keto – (55315) 3
+
+
Using the right-hand rule
vector
F = qvectorv ×
vector
B , the
top will be positive and the bottom will be
negative.
The other three choices are incorrect since
they do not abide by the right-hand rule.
2.
v
B
B
-
-
cor-
rect
008 10.0points
A thin, horizontal copper rod is 0 . 876585 m
long and has a mass of 79 . 4284 g.
3.
v
B
B
-
-
+
+
4.
v
B
B
+
+
-
-
Explanation:
Positive charges will move in the direction
of the magnetic force, while negative charges
move in the opposite direction.
Study the case where the magnetic field B
is pointed out of the plane of the paper and
the bar is moving from right to left ( ⇐ ).
To produce the indicated charge separa-
tion, the positive charges in the conductor
experience upward magnetic forces while the
negative charges in the conductor experience
downward magnetic forces leaving the charge
separation shown in the figure.
F F
v
B
B
+
+
-
-