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Andre45 [30]
4 years ago
12

A uniform magnetic field points upward, parallel to the page, and has a magnitude of 7.85 mt. a negatively charged particle (q=−

3.32 μc, m=2.05 pg) moves through this field with a speed of 67.3 km/s perpendicular to the magnetic field, as shown. the magnetic force on this particle is a centripetal force and causes the particle to move in a circular path. what is the radius of the particle's circular path?
Physics
1 answer:
Delvig [45]4 years ago
6 0
The Lorentz force exerted on the particle due to the magnetic field provides the centripetal force that keeps the particle in circular motion:
m \frac{v^2}{r} = qvB
where
m is the particle mass
v is its velocity
r its orbital radius
q its charge
B the magnetic field intensity
and where we neglected the factor \sin \theta in the Lorentz force formula because the particle is traveling perpendicular to the magnetic field, so \sin \theta=1 .

Re-arranging the formula, we get
r= \frac{mv}{qB} (1)
The problem gives us all the data about the particle and the magnetic field:
m=2.05 pg= 2.05 \cdot 10^{-12} g=2.05 \cdot 10^{-15} kg
v=67.3 km/s = 67.3 \cdot 10^3 m/s
q=3.32 \mu C = 3.32 \cdot 10^{-6} C (we are only interested in the magnitude of the charge)
B=7.85 mT = 7.85 \cdot 10^{-3}T

And by plugging these numbers into eq.(1), we find the radius of the orbit
r= \frac{mv}{qB}= \frac{(2.05 \cdot 10^{-15} kg)(67.3 \cdot 10^3 m/s)}{(3.32 \cdot 10^{-6} C)(7.85 \cdot 10^{-3} T)}=5.29 \cdot 10^{-3} m
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