contestada

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?

Respuesta :

skyluke89
The Lorentz force exerted on the particle due to the magnetic field provides the centripetal force that keeps the particle in circular motion:
[tex]m \frac{v^2}{r} = qvB[/tex]
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 [tex]\sin \theta[/tex] in the Lorentz force formula because the particle is traveling perpendicular to the magnetic field, so [tex]\sin \theta=1[/tex] .

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

And by plugging these numbers into eq.(1), we find the radius of the orbit
[tex]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 [/tex]

Otras preguntas

Q&A Education