Physics Physics 2102

Transcription

1 Physics 2102 Jonathan Dowling Physics 2102 Exam 2: Review Session Chapters / HW04-06 Some links on exam stress:

2 Exam 2 (Ch24) Sec.11 (Electric Potential Energy of a System of Point Charges); Sec.12 (Potential of Charged Isolated Conductor) (Ch 25) Capacitors: capacitance and capacitors; caps in parallel and in series, dielectrics; energy, field and potential in capacitors. (Ch 26) Current and Resistance: current, current density and drift velocity; resistance and resistivity; Ohm s law. (Ch 27) Circuits: emf devices, loop and junction rules; resistances in series and parallel; DC single and multiloop circuits, power; RC circuits. (Ch 28) Magnetic Fields: F=vxB, Right Hand Rule, Circular Motion, Force on Wire, Magnetic Dipole.

3 Potential V of Continuous Charge Distributions r = R! 2 + z 2 dv = kdq r V =! dv Straight Line Charge: dq=λdx λ=q/l Curved Line Charge: dq=λds λ=q/2πr Surface Charge: dq=σda σ=q/πr 2 da=2πr dr

4 Potential V of Continuous Charge Distributions Curved Line Charge: dq=λds λ=q/2πr Straight Line Charge: dq=λdx λ=q/l

5 Potential V of Continuous Charge Distributions Surface Charge: dq=σda σ=q/πr 2 da=r dr dϕ Straight Line Charge: dq=λdx λ=bx is given to you.

6 Capacitors E = σ/ε 0 = q/aε 0 E = V d q = C V C = ε 0 A/d Connected to Battery: V=Constant Disconnected: Q=Constant C = κ ε 0 A/d C=ε 0 ab/(b-a)

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

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

9 Resistors and 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

10 Capacitors and Resistors in Series and in Parallel What s the equivalent resistance (capacitance)? What s the current (charge) in each resistor (capacitor)? What s the potential across each resistor (capacitor)? What s the current (charge) delivered by the battery?

11 RC Circuits Time constant: RC Charging: q( t) = CE 1! e Discharging: q( t) =! t / RC ( ) q e 0 i(t)=dq/dt! t / RC

12 Capacitors: Checkpoints, Questions

13 Problem When switch S is thrown to the left, the plates of capacitor 1 acquire a potential V 0. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges q 1, q 2, and q 3 on the capacitors? Series

14 Series Seri-Q

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16 Current and Resistance: Checkpoints, Questions

17 Problem 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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19 Circuits: Checkpoints, Questions

20 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.

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22 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 10.0 µc?

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

25 Top view! = iab sin" net CF = 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! with B. The magnitude of the magnetic force on sides 1 and 3 is F = F = iab sin 90 = iab. The magnetic force on sides 2 and 4 is 1 3 F = F = ibb sin(90 "! ) = ibb cos!. 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!. The two 1 3 torques tend to rotate the loop in the same (clockwise) direction and thus add up. The net torque # = # 1+ # 3=( iabb / 2) sin! + ( iabb / 2) sin! = iabb sin! = iab sin!. (28-13) net

26 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! = NiAB.! We define a new vector µ associated with the coil, which is known as the magnetic dipole moment of the coil.!! = 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! = µ B sin " where " is the angle between µ and B.!!! In vector form: = µ # B.!! The potential energy of the coil is: U = $ µ B cos" = $ µ % B. U has a minimum value of $ µ B for " = 0 (position of stable equilibrium). U has a maximum value of µ B for " = 180 (position of unstable equilibrium). Note : For both positions the net torque is! = 0. (28-14)

27 Ch 28: Checkpoints and Questions

28 Circular Motion: Since magnetic force is perpendicular to motion, the movement of charges is circular. F v out F centrifugal = ma = mr! 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).

29 C. v! F! r.! B electron r = mv qb! = v r = qb m Radius of Circlcular Orbit Angular Frequency: Independent of v T! 2"r v = 2"mv qbv = 2"m qb Period of Orbit: Independent of v f! 1 T = qb 2"m Orbital Frequency: Independent of v

30 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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