- #1
fight_club_alum
- 63
- 1
- Homework Statement
- A charged particle (m = 5.0 g, q = –70 μC) moves horizontally at a constant speed of
30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s2 downward,
the electric field is 700 N/C upward, and the magnetic field is perpendicular to the
velocity of the particle. What is the magnitude of the magnetic field in this region?
a . 47 mT <-- answer
b. zero
c. 23 mT
d. 35 mT
e. 12 mT
- Relevant Equations
- F = ma
q . v . b . sin(theta) = F
m = 0.005
q = -70 x 10^-6 c
v = 30,000 m/s
Since there is no movement vertically Fb = Mg
So,
q . V . B = mg
So,
(70 x `10^-6) . (30,000) . B = (0.005) . (9.8)
So,
B = 0.0233333 or ~ 23 MT
q = -70 x 10^-6 c
v = 30,000 m/s
Since there is no movement vertically Fb = Mg
So,
q . V . B = mg
So,
(70 x `10^-6) . (30,000) . B = (0.005) . (9.8)
So,
B = 0.0233333 or ~ 23 MT