Physics1C_Chapter28 - 3/29/10 PHYSICS 1C LECTURE 1 Vahé...

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Unformatted text preview: 3/29/10 PHYSICS 1C LECTURE 1 Vahé Peroomian PHYSICS 1C ELECTRODYNAMICS, OPTICS, AND SPECIAL RELATIVITY   Mondays, Tuesdays, Wednesdays, and Fridays at 1:00 P.M. Kinsey Pavilion 1220B   Instructor:         Dr. Vahé Peroomian Office: 3860 Slichter Hall Phone: (310) 825-4114 E-mail: Text: University Physics, Young and Freedman, 12th Edition OFFICE HOURS   Mondays, Tuesdays and Wednesdays, 2:00 pm – 3:00 pm and by appointment Hours will be conducted in 3850 Slichter Hall   Office 1 3/29/10 TEACHING ASSISTANT   Lauren Pearce Email: Office Hours: TBA EXAMS   First Midterm   Monday, April 19, 2010 In class Second Midterm   Monday, May 17, 2010 In class   Final Exam   Thursday, June 10, 2010 11:30 am – 2:30 am   EXAM POLICIES   For both midterms and the final you will need your student I.D. One 4” × 6” sheet of notes, in your own handwriting, will be allowed during exams.   Makeup exams will not be given. If you miss a midterm, then the grading percentages of the other midterm and the final exam will be adjusted to 30% and 60%, respectively. This will only occur if and only if I receive a medical excuse letter signed by a physician. You cannot miss the final exam!!!!   Midterm exam grades will be in terms of points and percentages; letter grades will not be assigned. 2 3/29/10 HOMEWORK           Homework will be done/turned in through the Mastering Physics website, If you did not receive a Mastering Physics access code with your textbook, you can purchase one from the Mastering Physics website. You will need to enroll in my course, ID# PEROOMIAN1CS10 Each homework assignment will consist of a number of practice self-tutoring problems and 10 graded problems. Homework #1 is available now, and is due by Wednesday, April 7, 11:59pm. GRADING Homework Midterm #1   Midterm #2   Final Exam Total     10% 20% 20% 50% 100% CLASS WEBSITE (go to, click on “physics,” then “classes” and on “1C”)   Check class website frequently for:         Homework assignments PDFs of my lecture PowerPoints Sample exams, etc. Class website will be updated later today with HW #1. 3 3/29/10 HELPFUL HINTS         PDF of PowerPoints used in class posted to class website. Many of the examples in my lectures are not from our textbook. Make sure you do all the textbook examples!!! If you do homework in a group, please make sure you know how to do each problem individually. When studying for exams, don’t be satisfied with just redoing the problems I did in class plus homework. The more extra problems you do, the better. EXTRA HELP       Aside from my office hours and the TA’s office hours, extra help is available through the Physics 98 workshop program run by Brent Corbin. The workshop for this course meets Mondays and Wednesdays from 3:00 pm - 4:30 pm (starting week 2). To attend the workshop, you need to enroll in Physics 98XB Lab 3, a 1 unit P/NP course (with no impact on unit total), which will also give you access to the 98X website where you can get worksheets, solutions, good informational handouts, chatroom access, etc. A QUICK REVIEW OF SOME CHAPTER 27 CONCEPTS Vahé Peroomian 4 3/29/10 MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR   The magnetic force on a straight wire segment is given by: F= I l ×B   Magnetic force on infinitesimal wire section: dF = I d l × B EXAMPLE 27.7 A straight horizontal copper rod carries a current of 50.0 A from west to east in a region between the poles of a large electromagnet. In this region there is a horizontal magnetic field toward the north-east (45° north of east) with magnitude 1.20 T. (a) Find the magnitude and direction of the force on a 1.00 m section of rod. (b) While keeping the rod horizontal, how should it be oriented to maximize the magnitude of the force? What is the force magnitude in the case? EXAMPLE 1 A wire bent into a semicircle of radius R forms a closed circuit and carries a current I. The wire lies in the x-y plane, and a uniform magnetic field is directed along the positive y axis. Find the magnitude and direction of the magnetic force acting on the straight portion of the wire and on the curved portion. 5 3/29/10 FORCE AND TORQUE ON A CURRENT LOOP   The net force on a current loop in a uniform magnetic field is zero. However, the net torque is not in general equal to zero. THE MAGNETIC DIPOLE MOMENT   The magnetic dipole moment (µ) is the magnetic analog of the electric dipole moment (Chapter 21). For a loop of N turns, µ = NIA TORQUE AND POTENTIAL ENERGY   The vector torque on a current loop is given by τ = µ×B   Potential energy for a magnetic dipole is given by: U = − µ ⋅ B = − µ B cos φ 6 3/29/10 EXAMPLE 2 A rectangular coil of dimensions 5.40 cm × 8.50 cm consists of 25 turns of wire and carries a current of 15.0 mA. A 0.350-T magnetic field is applied parallel to the plane of the coil. (a) Calculate the magnitude of the magnetic dipole moment of the coil. (b) What is the magnitude of the torque acting on the loop? EXAMPLE 3 Consider the loop of wire in the figure. Imagine it is pivoted along side 4, which is parallel to the z axis and fastened so that side 4 remains fixed and the rest of the loop hangs vertically in the gravitational field of the Earth but can rotate around side 4 (panel b). The mass of the loop is 50.0 g, and the sides are of lengths a = 0.200 m and b = 0.100 m. The loop carries a current of 3.50 A and is immersed in a vertical uniform magnetic field of magnitude 0.010 0 T in the positive y direction (panel c). What angle does the plane of the loop make with the vertical? CHAPTER 28 SOURCES OF MAGNETIC FIELD Vahé Peroomian 7 3/29/10 HOW ARE MAGNETIC FIELDS CREATED   A point charge q moving with constant velocity v creates a vector magnetic field: µ qv × r ˆ B= 0 4π r 2 B= µ0 q v sin φ 4π r2 PERMEABILITY OF FREE SPACE   The constant µ0 is called the “permeability of free space,” and has the value µ0 = 4π × 10 −7 T ⋅ m/A   Relationship between µ0 and ε0: c2 = 1 ε 0 µ0 MAGNETIC FIELD OF A CURRENT ELEMENT   The total magnetic field caused by several moving charges is the vector sum of the fields caused by the individual charges µ0 Idl × r ˆ dB = 4π r 2   Biot-Savart Law µ B= 0 4π ∫ ˆ Idl × r 2 r 8 3/29/10 MAGNETIC FIELD OF A LONG STRAIGHT WIRE   The field near a long, straight currentcarrying conductor is given by B= µ0 I 4π a ∫ (x −a B= 2 x dy + y 2 ) 3/ 2 µ0 I 2π r EXAMPLE 1 Consider a thin, straight wire carrying a constant current I and placed along the x axis as shown in the figure. Determine the magnitude and direction of the magnetic field at point P due to this current. EXAMPLE 2 Consider a circular wire loop of radius a located in the y-z plane and carrying a steady current I as in the figure. Calculate the magnetic field at an axial point P a distance x from the center of the loop. 9 3/29/10 EXAMPLE 28.4 Two long straight parallel wires in the z direction, perpendicular to the x-y plane, each carry a current I in opposite directions. (a) Find the magnitude and direction of B at points P1, P2, and P3. (b) Find the magnitude and direction of B at any point on the x-axis on the right of wire 2 in terms of the x-coordinate of the point. FORCE BETWEEN PARALLEL CONDUCTORS   The force per unit length between two long, parallel, current-carrying conductors is given by F µ II ′ L = 0 2π r THE AMPERE   When the magnitude of the force per unit length between two long, parallel wires that carry identical currents and are separated by 1 meter is 2 × 10–7 N/m, the current in each wire is defined to be 1 Ampere. 10 3/29/10 EXAMPLE 4 Two infinitely long, parallel wires are lying on the ground a distance a = 1.00 cm apart as shown in the figure. A third wire, of length L = 10.0 m and mass 400 g, carries a current of I1 = 100 A and is levitated above the first two wires, at a horizontal position midway between them. The infinitely long wires carry equal currents I2 in the same direction, but in the direction opposite that in the levitated wire. What current must the infinitely long wires carry so that the three wires form an equilateral triangle? EXAMPLE: MAGNETIC FIELD OF A CIRCULAR CURRENT LOOP   Consider a circular wire loop of radius a located in the y-z plane and carrying a steady current I. Calculate the magnetic field at an axial point P a distance x from the center of the loop. MAGNETIC FIELD ON THE AXIS OF A COIL   What is the magnetic field along the axis of a coil with N circular loops? Bx = µ0 NIa 2 2 ( x 2 + a 2 ) 3/ 2 Bx = µ0 NI 2a 11 3/29/10 AMPERE’S LAW Analogous to Gauss’s Law for electric fields, evaluates the line integral of B around a closed path.   the Magnetic Field B.dl ∫ = µ0 I enclosed ere r , R . Here the current I 9 passing through the plane of circle 2 is less than circle 2 to t he area pr 2 l area pR 2 of the pr 2 Ir 5 I pR 2 ES S XAMPLE 5 r2 C B ? d s 5 B 1 2pr 2 5 m0I r 5 m0 a R 2 I b Ir 5 r2 I R2 A long, straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire. m0 I Calculate the magnetic field a distance r from the center of the for nd r ≤ ( B 5 ire in the2 regions(r ≥ Rr a, R) R. w a 2pR b r .) the wire is identiB t he case in highly Figure . (Example 30.5) use Ampère’s law Br Magnitude of the magnetic field .1). The magnetic B 1/r versus r for the wire shown in Figto the expression ure 30.13. The field is proportional harged sphere (see to r inside the wire and varies as 1/r r g netic field versus R outside the wire. re 30.14. Inside the ions 30.14 and 30.15 give the same value of the magnetic field at r 5 R , demonous at the surface of the wire. gnetic Field Created by a Toroid E en used to create an almostXAMPLE 28.9 area. The device consistsA solenoid consists of a helical winding of wire on a cylinder, of S S Loop 1 ds B (a torus) made of a nonconusually circular in cross-section. There can be hundreds or thousands of closely-spaced turns, losely spaced turns of wire, ea n occupied by the torus, ach of which can be regarded as a circular loop. Use Ampere’s Law to r find the field at or near the center of such a long solenoid. The solenoid b has n turns of wire per unit length and c a carries a current I. I lly to understand how the orus could be a solid matepped into the shape shown gh degree of symmetry, we aw problem. loop (loop 1) of radius r in t he magnitude of the field S S t, so B ? d s 5 B ds. Furthert imes, so the total current I Loop 2 Figure . (Example 30.6) A toroid consisting of many turns of wire. If the turns are closely spaced, the magnetic field in the interior of the toroid is tangent to the dashed circle (loop 1) and v aries as 1/r. T he dimension a is the cross-sectional r adius of the torus. The field outside the toroid is very small and can be described by using the amperian loop (loop 2) at the right side, perpendicular to the page. 12 3/29/10 EXAMPLE 6 A device called a toroid is often used to create an almost uniform magnetic field in some enclosed area. The device consists of a conducting wire wrapped around a ring (a torus) made of a nonconducting material. For a toroid having N closely spaced turns of wire, calculate the magnetic field in the region occupied by the torus, a distance r from the center. 13 ...
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This note was uploaded on 06/03/2011 for the course PHYSICS 1c taught by Professor Peroomian during the Spring '10 term at UCLA.

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