Lecture 32: MON 09 NOV Review Session A : Midterm 3

Transcription

1 Physics 2113 Jonathan Dowling Lecture 32: MON 09 NOV Review Session A : Midterm 3

2 EXAM 03: 6PM WED 11 NOV in Cox Auditorium The exam will cover: Ch.27.4 through Ch.30 The exam will be based on: HW08 11 The formula sheet and practice exams are here: Note this link also includes tutorial videos on the various right hand rules. You can see examples of even older exam IIIs here:

3 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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5 Magnetic Forces and Torques F = q v B + q E F v mv r = qb df = i dl B L τ = µ B

6 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

7 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)

8 Ch 28: Checkpoints and Questions

9 28.3: Finding the Magnetic Force on a Particle: v = index = forefinger B = "bird" finger F = thumb v F " = v " B " B " Always assume particle is POSITIVELY charged to work Out direction then flip your thumb over if it is NEGATIVE.

10 (a) + z F " = v " B " (b) x v B " (c) sin180 = 0

11 F 1 = 0 B + E F 3 = 0 B + E F 2 = B + E (a) F 2 > F 1 = F 3 F 4 = B + E F " = v " B " (b) F 4 = 0 v B "

12 ICPP The left face is at a lower electric potential (minus charges) and the right face is at a higher electric potential (plus charges).

13 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).

14 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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16 + (a) The electron because q = e for both particles but m p >> m e r p >> r e (b) Proton counter-clockwise and electron clockwise by right-hand-rule. r F v Which has the longer period T? T = 2πr v Since v is the same and r p >> r e the proton has the longer period T. It has to travel around a bigger circle but at the same speed. B into blackboard. r = mv qb

17 ICPP Two charged ions A and B traveling with a constant velocity v enter a box in which there is a uniform magnetic field directed out of the page. The subsequent paths are as shown. What can you conclude? (a) Both ions are negatively charged. A B v v (b) Ion A has a larger mass than B. (c) Ion A has a larger charge than B. (d) None of the above. r = mv qb RHR says (a) is false. Same charge q, speed v, and same B for both masses. So: ion with larger mass/charge ratio (m/q) moves in circle of larger radius. But that s all we know Don t know m or q separately.

18 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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22 Magnetic Force on a Wire. L F = i L B df = i dl B

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24 A portion of a loop of wire passes between the poles of a magnet as shown. We are viewing the circuit from above. When the switch is closed and a current passes through the circuit, what is the movement, if any, of the wire between the poles of the magnet? a) The wire moves toward the north pole of the magnet. b) The wire moves toward the south pole of the magnet. c) The wire moves upward (toward us). d) The wire moves downward (away from us into board). e) The wire doesn t move. i

25 π π Example F F = 1 = 3 ilb ibrdθ df = ibdl = F = sin( θ ) df = ibr sin( θ ) dθ 2iBR = By symmetry, F 2 will only have a vertical component, F = F + F + F = ilb + 2iRB + ilb = 2iB( L + total Notice that the force is the same as that for a straight wire of length R, L R R L Wire with current i. Magnetic field out of page. What is net force on wire? and this would be true no matter what the shape of the central segment. R)

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27 SP28-06 F grav = mg F mag = ilb m / L = kg/m To balance: F grav = F mag ICPP: Find Direction of B i mg = ilb B = kg m 9.8 m s 2 s 28 C B = mg il = m L g i = kg s C = T = 16 mt

28 Example 4: The Rail Gun Conducting projectile of length 2cm, mass 10g carries constant current 100A between two rails. Magnetic field B = 100T points outward. Assuming the projectile starts from rest at t = 0, what is its speed after a time t = 1s? rails B I L projectile Force on projectile: F= ilb (from F = il x B) Acceleration: a = F/m = ilb/m (from F = ma) v = at = ilbt/m (from v = v 0 + at) = (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s = 4,473mph = MACH 8

29 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

30 Torque on a Current Loop: Rectangular coil: A=ab, current = i Principle behind electric motors. Net force on current loop = 0 But: Net torque is NOT zero F F = = 3 iab 1 F = F sin( θ ) 1 Torque τ = F b = iabbsin(θ ) = For a coil with N turns, τ = N I A B sinθ, where A is the area of coil

31 Magnetic Dipole Moment We just showed: τ = NiABsinθ N = number of turns in coil A = area of coil. Define: magnetic dipole moment m µ, µ = ( NiA) nˆ nˆ τ = µ B Right hand rule: curl fingers in direction of current; thumb points along µ As in the case of electric dipoles, magnetic dipoles tend to align with the magnetic field.

32 Electric vs. Magnetic Dipoles p=qa +Q µ = ( NiA) nˆ -Q q QE QE τ = E p E U E = p E τ = µ B B U = µ B B

33 τ = µ B τ = µbsinθ 1 and 3 are downhill. 2 and 4 are uphill. U 1 = U 4 > U 2 = U 3 τ 1 = τ 2 = τ 3 = τ 4 τ is biggest when B is at right angles to μ

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

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

38 Superposition: ICPP Magnetic fields (like electric fields) can be superimposed -- just do a vector sum of B from different sources The figure shows four wires located at the 4 corners of a square. They carry equal currents in directions indicated What is the direction of B at the center of the square? I-OUT I-IN I-OUT I-IN B

39 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).