A 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. eptu) The magnitude of the magnetic field due to the straight segments at the center of the arc is:B.straight2. u) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet3. (4.) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculatedusing:10-7 ×2XTXX(6)(0.0344)10-7x2xxx(6)(0.0344)10-7x2xxx(6)(0.0344)10-7x2x*xx(6)(0.0344)4. (.3) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is:Barc5. (33) The direction of the magnetic field due to the three-quarter-circular arc at the center of the arc is pointing to:OInto the pageOOut of the pageOUpwardODownward

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A 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 Electron 1. ptu) The magnitude of the magnetic field due to the straight segments at the center of the arc is: B. straight 2.い ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet 3. (.. ) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated using: 10-7×2×オxニx(6) (0.0344) 10-7×2×オx-x(6) (0.0344) 10-7x2xxxx(6) (0.0344) 10-7x2xTxx(6) (0.0344) 4. (. 3) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is: Barc 5. (3 3) 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

A 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 Electron 1. ptu) The magnitude of the magnetic field due to the straight segments at the center of the arc is: B. straight 2.い ) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet 3. (.. ) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated using: 10-7×2×オxニx(6) (0.0344) 10-7×2×オx-x(6) (0.0344) 10-7x2xxxx(6) (0.0344) 10-7x2xTxx(6) (0.0344) 4. (. 3) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is: Barc 5. (3 3) 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

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Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

A 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. eptu) The magnitude of the magnetic field due to the straight segments at the center of the arc is:
B.
straight
2. u
) The magnetic field due to the arc can be calculated using the formula number (refer to the formula sheet
3. (4.
) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc can be calculated
using:
10-7 ×2XTXX(6)
(0.0344)
10-7x2xxx(6)
(0.0344)
10-7x2xxx(6)
(0.0344)
10-7x2x*xx(6)
(0.0344)
4. (.
3) The magnitude of the magnetic field due to the three-quarter-circular arc at the center of the arc is:
Barc
5. (3
3) 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
OOut of the page
OUpward
ODownward

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