Two long straight wires carrying 6 A of conventional current are connected by a three-quarter-circular arc of radius 0.0613 m. An electric field is Problem 9: ( also present in this region, due to charge not shown on the drawing. An electron is moving to the right with a speed of 4x 105 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: 9. =-1.6x 10-19 C)
Two long straight wires carrying 6 A of conventional current are connected by a three-quarter-circular arc of radius 0.0613 m. An electric field is Problem 9: ( also present in this region, due to charge not shown on the drawing. An electron is moving to the right with a speed of 4x 105 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: 9. =-1.6x 10-19 C)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:Problem 9: (
wires carrying 6 A of conventional current
are connected by a three-quarter-circular
arc of radius 0.0613 m. An electric field is
Two long straight
also present in this region, due to charge
not shown on the drawing. An electron is
moving to the right with a speed of 4x 105
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:
9 =-1.6x 10-19 C)
Radius R
C.
Electron
1. (.
magnetic field due to the straight
segments at the center of the arc is:
The magnitude of the
B.
%3D
straight
2.
arc can be calculated using the formula
number (refer to the formula sheet
The magnetic field due to the
) The magnitude of the magnetic
3. .
field due to the three-quarter-circular are
at the center of the arc can be calculated
using:
10-7x2xxx-x(6)
(0.0613)
10-"x2xxx-x(6)
(0.0613)
10-7x2xxx-x(6)
(0.0613)
10-"x2xrx-x(6)
(0.0613)
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