3. A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.20 × 10° N/C, while the magnetic field is 0.390 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.00 cm. Find the charge-to-mass ratio of the particle. C/kg

University Physics Volume 2
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Chapter11: Magnetic Forces And Fields
Section: Chapter Questions
Problem 65AP: An electron moving with a velocity v=(4.0i+3.0j+2.0k)106m/s enters a region where there is a uniform...
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3. A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity
selector is 3.20 × 103 N/C, while the magnetic field is 0.390 T. When the electric field is turned off, the charged particle travels on
a circular path whose radius is 4.00 cm. Find the charge-to-mass ratio of the particle.
C/kg
Transcribed Image Text:3. A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.20 × 103 N/C, while the magnetic field is 0.390 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.00 cm. Find the charge-to-mass ratio of the particle. C/kg
Expert Solution
Step 1

Electric and magnetic fields in a velocity selector are aligned perpendicular to each other so that only particles on which electric force is equal in magnitude and opposite in direction to the magnetic force could pass through the selector.

This particle must have a specific velocity ν. Otherwise particle will not pass through velocity selector.

F= FM

qE = qνB

νEB3.20 ×1030.390 = 8.205 ×103 m/s

 

 

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