-nints) Two long straight wires carrying 6 A of conventional current are connected by a three-quarter-circular arc of radius 0.0344 m. An electric field is also present in this region, due to charge not shown on the drawing. An electron is moving to the right with a speed of 5x 10 m/s as it passes through the center C of the arc, and at that instant the net electric and magnetic force on the electron is zero. (Consider the charge of the electron: q, = - 1.6x 10 19 C) Radius R C.- Electron 1.(. 1 ) The magnitude of the magnetic field due to the straight segments at the center of the arc is: %3! straight 2. ( ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet r) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated 3. (1 using 10-7x2xmxx(6) (0.0344) 10 7×2XTX-x (6) (0.0344) 10-7x2xx-x(6) (0.0344) 1/5 10-7 ×2×TX-x (6) (0.0344) 4. (* s) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is: Bare = 5. (. -) The direction of the magnetic field due to the three-quarter-circular arc at the center of the arc is pointing to: OInto the page COut of the page OUpward ODownward
-nints) Two long straight wires carrying 6 A of conventional current are connected by a three-quarter-circular arc of radius 0.0344 m. An electric field is also present in this region, due to charge not shown on the drawing. An electron is moving to the right with a speed of 5x 10 m/s as it passes through the center C of the arc, and at that instant the net electric and magnetic force on the electron is zero. (Consider the charge of the electron: q, = - 1.6x 10 19 C) Radius R C.- Electron 1.(. 1 ) The magnitude of the magnetic field due to the straight segments at the center of the arc is: %3! straight 2. ( ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet r) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated 3. (1 using 10-7x2xmxx(6) (0.0344) 10 7×2XTX-x (6) (0.0344) 10-7x2xx-x(6) (0.0344) 1/5 10-7 ×2×TX-x (6) (0.0344) 4. (* s) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is: Bare = 5. (. -) The direction of the magnetic field due to the three-quarter-circular arc at the center of the arc is pointing to: OInto the page COut of the page OUpward ODownward
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter22: Magnetic Forces And Magnetic Fields
Section: Chapter Questions
Problem 80P
Related questions
Question
![--ints) Two long straight wires carrying 6 A of conventional current are connected by a three-quarter-circular arc of
radius 0.0344 m. An electric field is also present in this region, due to charge not shown on the drawing. An electron is moving to
the right with a speed of 5x 10 m/s as it passes through the center C of the arc, and at that instant the net electric and magnetic
force on the electron is zero. (Consider the charge of the electron: q. = -1.6x 10-19 C)
Radius R
C.-
Electron
1.(.
1 ) The magnitude of the magnetic field due to the straight segments at the center of the arc is:
straight
2. ( ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet
3. (1
1) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated
using
(0.0344)
107×2XTX-x (6)
(0.0344)
10-7x2xx-x(6)
(0.0344)
1/5
10-7 ×2×TX-x (6)
(0.0344)
4. (*
s) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is:
Bere =
5. (-.
-) The direction of the magnetic field due to the three-quarter-circular arc at the center of the arc is pointing to:
OInto the page
COut of the page
OUpward
ODownward](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a92e8f7-5a5f-4d8e-bdf9-725c0d60a42c%2Fefbef9f2-8522-4453-a6d8-552821db872e%2Fjt1pd6a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:--ints) Two long straight wires carrying 6 A of conventional current are connected by a three-quarter-circular arc of
radius 0.0344 m. An electric field is also present in this region, due to charge not shown on the drawing. An electron is moving to
the right with a speed of 5x 10 m/s as it passes through the center C of the arc, and at that instant the net electric and magnetic
force on the electron is zero. (Consider the charge of the electron: q. = -1.6x 10-19 C)
Radius R
C.-
Electron
1.(.
1 ) The magnitude of the magnetic field due to the straight segments at the center of the arc is:
straight
2. ( ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet
3. (1
1) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated
using
(0.0344)
107×2XTX-x (6)
(0.0344)
10-7x2xx-x(6)
(0.0344)
1/5
10-7 ×2×TX-x (6)
(0.0344)
4. (*
s) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is:
Bere =
5. (-.
-) The direction of the magnetic field due to the three-quarter-circular arc at the center of the arc is pointing to:
OInto the page
COut of the page
OUpward
ODownward
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