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Physics 1402A/B Lecture Notes - Lecture 3: Drift Velocity, Net Force

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School
Western University
Department
Physics
Course
Physics 1402A/B
Professor
Jan Cami

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Document Summary

Lecture 3 notes - magnetic force on a straight wire. The goals of this lecture were to compute the magnetic force on a current-carrying wire. Since the current consists of moving charges, we know that a magnetic field will thus apply a magnetic force to them. It will also apply a force on the wire itself depending on it"s material. Well if we"re applying this to the context of a current in a wire, then the charged particle would be an electron moving in the direction of the wire itself at all points in the wire. Well if we apply that to a context of a of wire length of l, then the displacement across the wire would be v = l/t. If we chop that entire wire up into individual segments of wire with a infinitesimally small length dl. Then for each section of wire, this equation becomes vav = dl/dt. Well that is the velocity of the electrons

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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Related Questions

The long, straight wire in the figure has a current i = 1.07 Aflowing in it. A square loop with 10-cm sides and a resistance of0.0309 O is positioned 10 cm away from the wire. The loop is thenmoved in the positive x-direction with speed v = 15.09 cm/s.

a) Find the direction of the induced current in the loop. (Enter -1for clockwise, +1 for counterclockwise).
-1

b) Identify the directions of the magnetic forces acting on allsides of the square loop.
(You can enter x, -x, y, -y, z, -z, for each of the 4 sides. Thefirst answer is for the top part of the loop, the next for theright side, then the bottom, and finally the left side. Separatethe answers for each side by commas. Use no spaces. Example: If youthink that the force vectors on the top and bottom point in thenegative z-direction [=into the page], and the force vectors on theleft and right point in the positive y-direction, then enter:"-z,y,-z,y".)
y,x,-y,-x

c) Calculate the magnitude of the net force acting on the loop atthe instant it starts to move.

The long, straight wire in the figure has a current i = 1.59 Aflowing in it. A square loop with 10-cm sides and a resistance of0.0109 O is positioned 10 cm away from the wire. The loop is thenmoved in the positive x-direction with speed v = 17.55 cm/s

a) Find the direction of the induced current in the loop. (Enter -1for clockwise, +1 for counterclockwise)

b) Identify the directions of the magnetic forces acting on allsides of the square loop.
(You can enter x, -x, y, -y, z, -z, for each of the 4 sides. Thefirst answer is for the top part of the loop, the next for theright side, then the bottom, and finally the left side. Separatethe answers for each side by commas. Use no spaces. Example: If youthink that the force vectors on the top and bottom point in thenegative z-direction [=into the page], and the force vectors on theleft and right point in the positive y-direction, then enter:"-z,y,-z,y".)

c) Calculate the magnitude of the net force acting on the loop atthe instant it starts to move

http://img121.imageshack.us/i/p036figure.png/

The figure shows a long straight wire carrying a current of 10.0 Ato the right. At a certain instant, an electron is seen close tothe wire, at a distance d = 1.20 mm from the wire, moving with avelocity v = 2.25 x 106 m/s, in the direction shown. Draw thevector showing the force on the electron. Calculate the magnitudeof the force. Neglect the thickness of the wire. Of course the bluedot does not mean that the electron is that big. The electron hasno diameter to worry about. Look for data on the electron in yourbook, or on Google if you need to.

(The diagram is a black arrow pointed to the right with a linecoming down from the center labeled d then a blue dot at the end ofthat and a arrow coming from the blue dot pointing toward the rightlabeled v. these two lines with the blue dot make a 90 degreeangle)