As shown below, a charged particle is moving through a region that has a uniform E-field (the red lines) and a uniform 2.5 T B-field. As a result of these two fields and the resulting forces (ignore gravity), the particle has a constant velocity of 126 m/s. A.) Determine the directions of FE & FB on the particle. Direction of FE Choose direction Direction of FB Choose direction ✓ B.) Determine the magnitude of the E-field and the direction of the B-field. Assume that the B-field is perpendicular to the velocity. E = Direction of B= [Choose direction

Modern Physics
3rd Edition
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Chapter2: Relativity Ii
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Problem 4P: A charged particle moves along a straight line in a uniform electric field E with a speed v. If the...
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As shown below, a charged particle is moving through a region that has a uniform E-field (the red lines) and a uniform 2.5 T B-field. As a result of these two fields and the resulting forces (ignore gravity), the
particle has a constant velocity of 126 m/s.
i
A.) Determine the directions of FE & FB on the particle.
Direction of FE
Direction of FB =
B.) Determine the magnitude of the E-field and the direction of the B-field. Assume that the B-field is perpendicular to the
velocity.
E =
Direction of B Choose direction
Choose direction ✓
Choose direction
Transcribed Image Text:As shown below, a charged particle is moving through a region that has a uniform E-field (the red lines) and a uniform 2.5 T B-field. As a result of these two fields and the resulting forces (ignore gravity), the particle has a constant velocity of 126 m/s. i A.) Determine the directions of FE & FB on the particle. Direction of FE Direction of FB = B.) Determine the magnitude of the E-field and the direction of the B-field. Assume that the B-field is perpendicular to the velocity. E = Direction of B Choose direction Choose direction ✓ Choose direction
Expert Solution
Step 1

constant velocity (v) = 126 msmagnitude of electric field = Emagnitude of magnetic field (B)= 2.5 Tsign of charge = positive 

 

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