DOC PREVIEW
UT Arlington PHYS 1444 - Lecture Notes

This preview shows page 1-2-3-4 out of 13 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

PHYS 1444 – Section 004 Lecture #14Magnetic Forces on Electric CurrentExample 27 – 1Example 27 – 2Example 27 – 3Charged Particle’s Path in Magnetic FieldExample 27 – 4Cyclotron FrequencyTorque on a Current LoopSlide 10Magnetic Dipole MomentMagnetic Dipole Potential EnergyExample 27 – 8Wednesday, Mar. 28, 2007 1 PHYS 1444-004, Spring 2007Dr. Andrew BrandtPHYS 1444 – Section 004Lecture #14Wednesday, Mar. 28 2007Dr. Andrew BrandtHW6 due Fri 3/30 at 8 pmInterim Grades posted•Magnetic Forces•TorqueWednesday, Mar. 28, 2007 2 PHYS 1444-004, Spring 2007Dr. Andrew Brandt Magnetic Forces on Electric Current•Direction of the force is always –perpendicular to the direction of the current and also–perpendicular to the direction of the magnetic field, B•Experimentally the direction of the force is given by another right-hand rule  When the fingers of the right-hand point to the direction of the current and the finger tips bent to the direction of magnetic field B, the direction of thumb points to the direction of the force•I prefer the fingers along the B field, the thumb along the current (same as in the direction of B field around a wire) and palm for the force.Wednesday, Mar. 28, 2007 3 PHYS 1444-004, Spring 2007Dr. Andrew BrandtExample 27 – 1 Measuring a magnetic field. A rectangular loop of wire hangs vertically as shown in the figure. A magnetic field B is directed horizontally perpendicular to the wire, and points out of the page. The magnetic field B is very nearly uniform along the horizontal portion of wire ab (length l=10.0cm) which is near the center of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward force ( in addition to the gravitational force) of F=3.48x10-2N when the wire carries a current I=0.245A. What is the magnitude of the magnetic field B at the center of the magnet? Magnetic force exerted on the wire due to the uniform field is Since Magnitude of the force is B l^rrF Il B= �rr rF IlB=Solving for BB =FIl=23.48 100.245 0.10NA m-�=�1.42TWhat happened to the forces on the loop on the side? The two forces cancel out since they are in opposite direction with the same magnitude.Wednesday, Mar. 28, 2007 4 PHYS 1444-004, Spring 2007Dr. Andrew BrandtExample 27 – 2 Magnetic force on a semi-circular wire. A rigid wire, carrying the current I, consists of a semicircle of radius R and two straight portions as shown in the figure. The wire lies in a plane perpendicular to the uniform magnetic field B0. The straight portions each have length l within the field. Determine the net force on the wire due to the magnetic field B0. As in the previous example, the forces on the straight sections of the wire are equal and in opposite directions. Thus they cancel. Since What is the net x-component of the force exerted on the circular section?0B dl^rrdF Idl B= �rr rdF =Integrating over F =What do we use to figure out the net force on the semicircle? We divide the semicircle into infinitesimal straight sections. dl Rdf=0 Why? Because the forces on left and the right-hand sides of the semicircle balance.y-component of the force dF is ( )sind F f =0IRB dfWhich direction? Vertically upward direction. The wire will be pulled deeper into the field. ( )0sind Fpf =�00sinIB R dpf =�[ ]00cosIB Rpf- =02RIBWednesday, Mar. 28, 2007 5 PHYS 1444-004, Spring 2007Dr. Andrew BrandtExample 27 – 3 Magnetic force on a proton. A proton having a speed of 5x106m/s in a magnetic field feels a force of F=8.0x10-14N toward the west when it moves vertically upward. When moving horizontally in a northerly direction, it feels zero force. What is the magnitude and direction of the magnetic field in this region? What is the charge of a proton? What does the fact that the proton does not feel any force in a northerly direction tell you about the magnetic field?Why? pq =The field is along the north-south direction. Because the particle does not feel any magnetic force when it is moving along the direction of the field. Since the particle feels force toward the west, the field should be pointing to …. North Using the formula for the magnitude of the field B, we obtainB =Fqv=1419 68.0 100.101.6 10 5.0 10 /NTC m s--�=״�We can use magnetic field to measure the momentum of a particle. How? 191.6 10e C-+ = �Wednesday, Mar. 28, 2007 6 PHYS 1444-004, Spring 2007Dr. Andrew Brandt•What is the shape of the path of a charged particle moving in a plane perpendicular to a uniform magnetic field?–Circle!! Why?–An electron moving to right at the point P in the figure will be pulled downwardCharged Particle’s Path in Magnetic Field–At a later time, the force is still perpendicular to the velocity–Since the force is always perpendicular to the velocity, the magnitude of the velocity is constant–The direction of the force follows the right-hand-rule and is perpendicular to the direction of the magnetic field–Thus, the electron moves in a circular path with a centripetal force F.Wednesday, Mar. 28, 2007 7 PHYS 1444-004, Spring 2007Dr. Andrew BrandtExample 27 – 4 Electron’s path in a uniform magnetic field. An electron travels at a speed of 2.0x107m/s in a plane perpendicular to a 0.010-T magnetic field. Describe its path. What is the formula for the centripetal force? Since the magnetic field is perpendicular to the motion of the electron, the magnitude of the magnetic force isSince the magnetic force provides the centripetal force, we can establish an equation with the two forces F =( ) ( )( )( )31 72199.1 10 2.0 101.1 101.6 10 0.010kg m smC T---״�= �� �F =F =r =Solving for rma =m2vrevBevB =2vmrmveB=Wednesday, Mar. 28, 2007 8 PHYS 1444-004, Spring 2007Dr. Andrew Brandt•The time required for a particle of charge q moving w/ constant speed v to make one circular revolution in a uniform magnetic field, , is•Since T is the period of rotation, the frequency of the rotation is•This is the cyclotron frequency, the frequency of a particle with charge q in a cyclotron accelerator–While r depends on v, the frequency is independent of v and r. Cyclotron FrequencyT =B v^rr12qBfT mp= =2 rvp=2vp2 mqBpmvqB=Wednesday, Mar. 28, 2007 9 PHYS 1444-004, Spring 2007Dr. Andrew Brandt–The magnetic field exerts a force on both vertical sections of wire.–Where is this principle used?•Ammeters, motors, volt-meters,


View Full Document

UT Arlington PHYS 1444 - Lecture Notes

Documents in this Course
Generator

Generator

14 pages

Load more
Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?