lecture 32: mon 09 nov review session a : midterm 3 physics 2113 jonathan dowling

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

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

Lecture 32: MON 09 NOVReview Session A : Midterm

3

Physics 2113

Jonathan Dowling

Page 2: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

EXAM 03: 6PM WED 11 NOV in Cox Auditorium

The exam will cover: Ch.27.4 through Ch.30The exam will be based on: HW08–11

The formula sheet and practice exams are here:http://www.phys.lsu.edu/~jdowling/PHYS21133-FA15/lectures/index.html

Note this link also includes tutorial videos on the various right hand rules.

You can see examples of even older exam IIIs here:http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys2102/Phys2102OldTests/

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

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?

Page 4: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 5: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Magnetic Forces and TorquesMagnetic Forces and Torques

L

vF

Page 6: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

C

C

Top viewSide view

(28-13)

Page 7: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

(28-14)

Page 8: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Ch 28: Checkpoints and QuestionsCh 28: Checkpoints and Questions

Page 9: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

28.3: Finding the Magnetic Force on a Particle:

Always assume particle is POSITIVELY charged to workOut direction then flip your thumb over if it is NEGATIVE.

Page 10: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 11: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 12: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

The left face is ata lower electric potential (minus charges) and the right face is at a higherelectric potential (plus charges).

ICPP

Page 13: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Circular Motion: Since magnetic force is perpendicular to motion, the movement of charges is circular.

B into blackboard.

v

F

In general, path is a helix (component of v parallel to field is unchanged).

In general, path is a helix (component of v parallel to field is unchanged).

r

Page 14: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

.

.electron

C

r

Radius of Circlcular OrbitRadius of Circlcular Orbit

Angular Frequency:Independent of v Angular Frequency:Independent of v

Period of Orbit:Independent of v Period of Orbit:Independent of v

Orbital Frequency:Independent of v Orbital Frequency:Independent of v

Page 15: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 16: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

B into blackboard.

v

F

r

Which has the longer period T?

Since v is the same and rp >> re the proton has the longer period T. It has to travel around a bigger circle but at the same speed.

Page 17: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

ICPPICPPTwo 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?

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.

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.

(a) Both ions are negatively

charged.

(b) Ion A has a larger mass than B.

(c) Ion A has a larger charge than B.

(d) None of the above.

v

v

A

B

Page 18: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

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?

Page 19: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 20: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 21: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 22: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Magnetic Force on a Wire.Magnetic Force on a Wire.

L

Page 23: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 24: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

i

28.8.2. 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.

Page 25: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

ExampleExample

By symmetry, F2 will only have a vertical component,

Notice that the force is the same as that for a straight wire of length R,

L LR R

and this would be true nomatter what the shape of the central segment!.

Wire with current i.Magnetic field out of page.What is net force on wire?

Page 26: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

i

Page 27: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

i

SP28-06

ICPP: Find Direction of B

Page 28: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Example 4: The Rail GunExample 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?

• 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?

B I L

• 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 = v0 + at)

= (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s= 4,473mph = MACH 8!

• 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 = v0 + at)

= (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s= 4,473mph = MACH 8!

projectile

rails

Page 29: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Highest Torque: θ = ±90° sinθ = ±1

Lowest Torque: θ= 0° & 180° sinθ = 0

Bθ = 180°–cosθ = +1

θ = 0°–cosθ = –1

Page 30: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Torque on a Current Loop: Principle behind electric motors.

Net force on current loop = 0

For a coil with N turns,τ = N I A B sinθ, where A is the area of coil

Rectangular coil: A=ab, current = i

But: Net torque is NOT zero!

Page 31: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Magnetic Dipole MomentMagnetic Dipole Moment

N = number of turns in coilA = area of coil.

We just showed: τ = NiABsinθRight hand rule:

curl fingers in direction of current;

thumb points along μDefine: magnetic dipole moment μ

As in the case of electric dipoles, magnetic dipoles tend to align with the magnetic field.

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

Electric vs. Magnetic DipolesElectric vs. Magnetic Dipoles

-Q

θQE

QE

+Q

p=Qa

Page 33: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

τ is biggest when B is at right angles to μ

1 and 3 are “downhill”.2 and 4 are “uphill”.U1 = U4 > U2 = U3

Page 34: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 35: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling
Page 36: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Right Hand Rule: Given Current i Find Magnetic Field B 

Page 37: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Checkpoints/QuestionsCheckpoints/QuestionsMagnetic field?

Force on each wire due to currents in the other wires?

Ampere’s Law: Find Magnitude of ∫B∙ds?

Page 38: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

Superposition: ICPPSuperposition: 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?

B

I-OUT I-OUT

I-IN I-IN

Page 39: Lecture 32: MON 09 NOV Review Session A : Midterm 3 Physics 2113 Jonathan Dowling

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).

Right Hand Rule: Given Current i Find Magnetic Field B