Lecture 31: MON 30 MAR Review Session : Midterm 3

Transcription

1 Physics 2113 Jonathan Dowling Lecture 31: MON 30 MAR Review Session : Midterm 3

2 EXAM 03: 8PM MON 30 MAR in Cox Auditorium The exam will cover: Ch.26 through Ch.29 The exam will be based on: HW07 HW10. The formula sheet for the exam can be found here: You can see examples of old exam IIIs here:

3 Current and Resistance i = dq/dt Junction rule R = ρl/a r = ρ 0 (1+a(T-T 0 )) V = i R E = J ρ

4 Current and Resistance

5 Current and Resistance: Checkpoints, Questions

6 A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5x10-5 Ωm. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?

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8 i = dq/dt Junction rule Circuits

9 DC Circuits Loop rule V = ir P = iv Single loop Multiloop

10 Resistors vs Capacitors Resistors Capacitors Key formula: V=iR Q=CV In series: same current R eq = R j same charge 1/C eq = 1/C j In parallel: same voltage same voltage 1/R eq = 1/R j C eq = C j

11 Resistors in Series and in Parallel What s the equivalent resistance? What s the current in each resistor? What s the potential across each resistor? What s the current delivered by the battery? What s the power dissipated by each resisitor?

12 Series and Parallel

13 Series and Parallel

14 Series and Parallel

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16 Circuits: Checkpoints, Questions

17 Problem: 27.P.018. [406649] Figure shows five 5.00 resistors. (Hint: For each pair of points, imagine that a battery is connected across the pair.) Fig (a) Find the equivalent resistance between points F and H. (b) Find the equivalent resistance between points F and G. Slide Rules: You may bend the wires but not break them. You may slide any circuit element along a wire so long as you don t slide it past a three (or more) point junction or another circuit element.

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19 Problem: 27.P.046. [406629] In an RC series circuit, E = 17.0 V, R = 1.50 MΩ, and C = 1.80 µf. (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 16.0 µc?

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21 Magnetic Forces and Torques!!!! F = q v B + q E F v mv r = qb!!! df = i dl B L! τ! = µ B!

22 Top view τ net = iab sinθ C F = net 0 Side view C Magnetic Torque on a Current Loop Consider the rectangular loop in fig. a with sides of lengths a and b and that carries a current i. The loop is placed in a magnetic field so that the normal nˆ to the loop! forms an angle θ with B. The magnitude of the magnetic force on sides 1 and 3 is F = F = iabsin 90 = iab. The magnetic force on sides 2 and 4 is 1 3 F = F = ibbsin(90 θ ) = ibb cos θ. These forces cancel in pairs and thus F = The torque about the loop center C of F and F is zero because both forces pass 2 4 through point C. The moment arm for F and F is equal to ( b / 2)sin θ. The two 1 3 torques tend to rotate the loop in the same (clockwise) direction and thus add up. The net torque τ = τ1+ τ 3=( iabb / 2) sin θ + ( iabb / 2) sinθ = iabb sinθ = iab sin θ. (28-13) net

23 U = µ B U = µ B Magnetic Dipole Moment The torque of a coil that has N loops exerted U by a uniform magnetic field B and carries a current i is given by the equation τ = NiAB.! We define a new vector µ associated with the coil, which is known as the magnetic dipole moment of the coil.!! τ = µ B!! = µ B! The magnitude of the magnetic dipole moment is µ = NiA. Its direction is perpendicular to the plane of the coil.! The sense of µ is defined by the right-hand rule. We curl the fingers of the right hand in the direction of the current. The thumb gives us the sense. The torque can be!! expressed in the form τ = µ Bsin θ where θ is the angle between µ and B.!!! In vector form: τ = µ B.!! The potential energy of the coil is: U = µ B cosθ = µ B. U has a minimum value of µ B for θ = 0 (position of stable equilibrium). U has a maximum value of µ B for θ = 180 (position of unstable equilibrium). Note : For both positions the net torque is τ = 0. (28-14)

24 Ch 28: Checkpoints and Questions

25 Circular Motion: F v Since magnetic force is perpendicular to motion, the movement of charges is circular. out F centrifugal = ma = mrω 2 = m v 2 r r B into blackboard. in F magnetic = qvb F B = F C qv B = mv 2 r Solve : r = mv qb In general, path is a helix (component of v parallel to field is unchanged).

26 C. v! F! r.! B electron r = mv qb ω = v r = qb m Radius of Circlcular Orbit Angular Frequency: Independent of v T 2πr v = 2πmv qbv = 2πm qb Period of Orbit: Independent of v f 1 T = qb 2πm Orbital Frequency: Independent of v

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28 Problem: 28.P.024. [566302] In the figure below, a charged particle moves into a region of uniform magnetic field, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 160 ns in the region. (a) What is the magnitude of B? (b) If the particle is sent back through the magnetic field (along the same initial path) but with 3.00 times its previous kinetic energy, how much time does it spend in the field during this trip?

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30 ! F = ilbsinθ! τ = µbsinθ = µbcosθ! µ B! µ Highest Torque: θ = ±90 sinθ = ±1 Lowest Torque: θ= 0 & 180 sinθ = 0 θ = 180 cosθ = +1 θ = 0 cosθ = 1

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35 Right Hand Rule: Given Current i Find Magnetic Field B

36 Checkpoints/Questions Magnetic field? Force on each wire due to currents in the other wires? Ampere s Law: Find Magnitude of B ds?

37 Right Hand Rule: Given Current i Find Magnetic Field B A C y l P x z B D +z is out of page B B B D B B,C B A B A B B,C B D B D B A B C B D BD B B,C B A BA B B,C The current in wires A,B,D is out of the page, current in C is into the page. Each wire produces a circular field line going through P, and the direction of the magnetic field for each is given by the right hand rule. So, the circles centers in A,B,D are counterclockwise, the circle centered at C is clockwise. When you draw the arrows at the point P, the fields from B and C are pointing in the same direction (up and left).

38 Right Hand Rule & Biot-Savart: Given i Find B A length of wire is formed into a closed circuit with radii a and b, as shown in the Figure, and carries a current i. (a) What are the magnitude and direction of B at point P? (b) Find the magnetic dipole moment of the circuit. B = µ 0iφ 4πR! B = µ=nia

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