--ints) Two long straight wires carrying 6 A of conventional current are connected by a three-quarter-circular arc ofradius 0.0344 m. An electric field is also present in this region, due to charge not shown on the drawing. An electron is moving tothe 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 magneticforce on the electron is zero. (Consider the charge of the electron: q, = - 1.6x 10 19 C)Radius RC.-Electron1. (.1 ) The magnitude of the magnetic field due to the straight segments at the center of the arc is:B.%3!straight2. ( ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet3. (1r) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculatedusing10-7x2xmxx(6)(0.0344)10 7x2xTx-x (6)(0.0344)10-7x2xx-x(6)(0.0344)1/510-7 ×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 pageCOut of the pageOUpwardODownward

Answered: Image /qna-images/answer/49160158-9caf-4ac2-b2a3-9bab82bca919.jpg

--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 1.6 ) The magnitude of the magnetic field due to the straight segments at the center of the arc is: B. %3D straight 2. (. ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet 3. (1 r) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated using 10x2xxxx(6) (0.0344) 10 7×2×TX-x (6) (0.0344) 10x2xx (0.0344) 10-7x2xxx(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: CInto the page COut of the page OUpward ODownward

--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 1.6 ) The magnitude of the magnetic field due to the straight segments at the center of the arc is: B. %3D straight 2. (. ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet 3. (1 r) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated using 10x2xxxx(6) (0.0344) 10 7×2×TX-x (6) (0.0344) 10x2xx (0.0344) 10-7x2xxx(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: CInto 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 51P

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Similar questions

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