A point charge q=+5.00mC moves through the origin along the x-axis with
velocity v. A uniform magnetic field B lies in the x-y.
The vectors are given by
v=4.20×105 m/s
^
i
B=(1.20 T)
^
i
+(2.40 T)
^
j
(a) Find the instantaneous force acting on the point charge as it moves
through the origin.
F
=
qv×B
=
(+5.00×10-6 C)
æ è
4.20×105 m/s
^
i
ö ø
×
é ë
æ è
1.20 T
^
i
ö ø
+
æ è
2.40 T
^
j
ö ø
ù û
=
(5.00×10-6 C)(4.20×105 m/s)(2.40 T)
æ è
^
i
×
^
j
ö ø
=
+5.04 N
^
k
(b) Show that F·v=0.
F·v
=
(5.04 N
^
k
)·
æ è
4.20×105 m/s
^
i
ö ø
=
2.12×106 N·m/s )
æ è
^
k
·
^
i
ö ø
=
0
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version 3.85. On 10 Mar 2016, 17:27.