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Physics 202 Exam 2 Review

About Exam 2When and where

Friday March. 13th 5:30-7:00 pm2650 3650 H manities (same as e am 1 )2650, 3650 Humanities (same as exam 1 )

FormatClosed book 1+1 8x11 formula sheet allowed, must be self prepared, no photo copy of solutions, no photo copy of lecture slides.Four full problems. (~20 questions)Bring a calculator (but no computer). Only basic calculation functionality can be used.

Special Arrangements: Spec a a ge e tsAll alternative time tests should have been preapproved.No early test before Friday Mar 13th possible.Alternative test available in earlier afternoon ( 4pm) in labAlternative test available in earlier afternoon (~ 4pm) in lab rooms (for approved requests only)

Chapters CoveredChapter 27: Electric CurrentsChapter 28: DC CircuitCh t 29 M tiChapter 29: Magnetism Chapter 30: Source of Magnetic Field

Disclaimer

This review is a supplement to your own preparationThis review is a supplement to your own preparation.

Hints and exercises presented in this review are not t t b l tmeant to be complete.

Exam Topics (1)Key concepts

(“key”: those in summary box at the end of each chapter)

Basic Quantities: Electrical Current (I), Voltage (ΔV)Resistance (R), resistivity (ρ)Power Consumed by RemfemfTime Constant, RC Magnetic ForceMagnetic Field Magnetic Field Lines Magnetic FluxMagnetic Field, Magnetic Field Lines, Magnetic FluxMagnetic Dipole Moment.

• (Definition, force, torque, potential energy)Permeability/susceptibility for ferro/para/dia magnetic materials.

Exam Topics(2)Current and resistance.

I= ΔQ/ΔtO f OOhm’s Law ΔV=IR for Ohmic materialsResistors in series and parallelPower consumption on RPower consumption on R

DC circuitKirchhoff’s Rules

• Junction rule• Loop rule

Simple 1-loop, 2-loop circuit of R’s and ε’s

Time constant and RC circuit

Exam Topics(3)Magnetic Force

Magnetic force has a form of qvxB.• always perpendicular to v and B• always perpendicular to v and B.• never does work• charged particle moves in circular/helix path in uniform

B field (ω=qB/m, r=mv/qB)• On current segment, it has the form ILxB

Uniform B closed loop ΣF=0 Στ=μxB– Uniform B, closed loop ΣF=0, Στ=μxBMagnetic Field:

Field lines, “north” and “south”. B field never does work.

M ti Di l M tF =∑ 0

Magnetic Dipole Moments :

B-UB•=

×=μ

μτr

vr

Exam Topics(4)Magnetic Fields can be produced by:

moving charge (Biot-Savart law)change of E field (displacement current, not in this exam.)

Ampere’s LawAmpere s LawAmpere’s law simplifies the calculation of B field in some symmetric cases.

• (infinite) straight line, (infinite) current sheet, Solenoid, Toroid

Gauss’s Law in Magnetism no magnetic charge.g g gForces between two currents

Can be attractive/repulsive

Reminder: Basic Current, Resistance, Power

Current I = ΔQ/Δt through a cross-section.

Resistance R l

Basic Current, Resistance, Power

Resistance:

Ohms Law: ΔV =RI

AR ρ=

General Electric Power : P= IΔVOhmic Electric Power: P= I2R = ΔV2/R

R1, R2 in series:I1=Ι2, ΔV1+ΔV2=ΔV R=R1+R2

R1, R2 in parallel:I=I1+Ι2, ΔV1=ΔV2=ΔV 1/R=1/R1+1/R2

Exercise 1: Two Light BulbsLight bulb A is rated at 12W when operated at 12V, light bulb B is rated at 3W when under 12V. Both bulbs are of resistive type (incandescent).(incandescent).What are their resistance RA and RB?

Assume the brightness of a bulb is proportional to its powerAssume the brightness of a bulb is proportional to its powerconsumption

When they are connected to a power source in parallel, which one is brighter?

When they are connected to a power source in series, which is brighter?

Reminder:Procedure to Use Kirchoff RulesProcedure to Use Kirchoff Rules

1. Assign a directional current for each branch (segment) of a circuit. The assigned direction for each current can be arbitrarily chosen but, once assigned, need to be observed.

1. Set up junction rules (for as many junctions as necessary: ΣIin =ΣIoutin out

3. Set up loop rules (as many as necessary): ΣΔV =0 p p ( y y)

3. Solve for unknowns. 3 So e o u o s

4. If a current is found to be negative, it means its actual direction is opposite to the originally chosen one. The pp g ymagnitude is always correct.

Determine Potential DifferencePut this on your formula sheet (no photo copy! ) y ( p py )

a b a ba bloop

direction

a bloop

direction

ΔV=Vb-Va= - ε ΔV=Vb-Va= +ε

I I

a bloop

direction

a bloop

directiondirection

ΔV=Vb-Va= -IR

direction

ΔV=Vb-Va= +IR

Exercise 2: A Circuit with Three emf’sIn the circuit shown, R1=1Ω, R2=2Ω, ε1= 3V, ε2= 1V, ε3= 2V. Use Kirchhoff’s rules to find the currents (magnitude and direction)

passing resistor R and Rpassing resistor R1 and R2.

Solution (see board). ε3

Answers:I1= 1A to the left, I2=0.5A to the right

R2

e2

R2

ε2

I3

I2

What is the total power consumed in circuit. R1

e1

R1

ε1

I2

I1

(Trivial once you have current, do it after class yourself)

Reminder: Charging A Capacitor in RC CircuitCharging A Capacitor in RC Circuit

)1()( / RCteCtq −−= ε RCteR

tI /)( −=ε

ChargingC Chargingτ≡RC :time constant

Reminder : Time Constant When DischargingTime Constant When Discharging

RCtQetq /)( −=

RCteQtI /)( −= eRC

tI )( −=

discharging

Again time constant τ=RC Everything determined by t/τ

Exercise 3: Time Constant and RCHW 6 problem 2 (Ch 28 P032)

R1

Cε S

R2

Cε S

Find out time constant when S openFind out time constant when S openFind out time constant when S closedTime dependency after S closed.

Ask your TA for help if in puzzle.

Reminder: Forces on Charges and CurrentForces on Charges and Current

On charged particle:F= qE + q vxB

On current segment:F= ILxB

Current inside uniform B fieldF=IL’ xB L’

I

B

L

dsds

Exercise 4: Motion of Charged Particle In Uniform B Field

At t=0, an electron of velocity v= vx i + vy j enters a uniform B field B k . (vx, vy >0)

What is the magnetic force on the electron at t=0What is the magnetic force on the electron at t=0.Convince yourself that the electron will go in a circular path once enters B,

What is the radius r?• What is the radius r?• On x-y plane, draw the path of the electron and indicate

its direction in the circle.

Solution: See board.j

Solution:F=-eVxB

= -e (vx i + vy j ) x B k= eB (v i x k + v j x k )= -eB (vx i x k + vy j x k )= -eB (-vx j+ vy i )

r= mv/(eB) what v to use?v

i

k: out of page

r= mv/(eB), what v to use? v=sqrt(vx

2+vy2) !

Exercise 5: Mass/Velocity SelectorThe schematic of a mass selector is shown below.For given E and B1, B2 ,

What is the speed of a particle of charge q that can enter fieldWhat is the speed of a particle of charge q that can enter field B2 ? Once inside field B2 , the above particle moves in a semicircle of radius r as shown what is the mass of the particle?of radius r as shown, what is the mass of the particle? and what is the sign of its charge?

B1Solution:While inside B1 and E:F=qE+qVxB1 = 0 v= E/B1

Once into B2:

m=(qB2r)/v = qB1B2r/E

B2

(q 2 ) q 1 2

Exercise 6:Magnetic Force on Seven Sides of an Current Carrying Octagon y g g

A wire carrying current I is placed in a uniform magnetic field B as shown. The part of the wire inside the magnetic field occupies 7 sides of an octagon of side length “a”. Find the totaloccupies 7 sides of an octagon of side length a . Find the total magnetic force on the current carrying wire in the figure.

Solution (see board)Solution (see board)

Answer: F= IaB direction: up.

Reminder: All Those Right-Hand Rules

Trivial Exercise: Solenoid and Bar Magnet A current carrying solenoid is placed near a bar magnet as shown, are they attractive or repulsive to

h th ?each other?

N SN S

Answer: Attractive

A k TA f i i i iAsk your TA for more imaginative exercises

Reminder:Two Ways to Calculate Magnetic Field

Biot-Savart Law (first principle): dB

∫×

= 20 ˆ

4 rdI rsB

πμr

Ampere’s Law:

4 rπdB’

I(Practical only for settings that are highly symmetric)

ds∫ =•any

IdsB 0μr

pathclosedany

Exercise 7: Biot-Savart LawUse Biot-Savart to find the magnetic field at the point P.

Solutions: (See board)

1∫

×= 2

0 ˆ4

dI rsB μr

Answer:2

∫ 24 rπ

Answer:

segment 1 contribution:

B 03

B=0

segment 3 contribution: B=0P

IdI ˆsegment 2:

(into page) RI

RdI

8ˆ

40

20 μπ

μ=

×= ∫

rsBr

Exercise 8: Ampere’s Law An infinite straight thin wire is at the center of two concentric conducting cylinders of radius R and 2R.

The currents are I (into the page), 2I (out ), and I (in) , respectivelyThe currents are I (into the page), 2I (out ), and I (in) , respectively for the center wire and the two cylinders. (as color coded).

Find B as function of r.

Solution:

IrBd enclosedμπ2 0==•∫ sBr

Answers:

rIB enclosed

πμ

20=⇒

Answers:

r<R , B= μ0I/2πr (Clockwise)

R<r<2R B= μ0I/2πr (counter-clockwise)R<r<2R , B= μ0I/2πr (counter-clockwise)

r>2R , B= 0

Field Test -- Exercise 9 (Part A) An infinite straight current I is place at (x,y)= (0,0) and runs perpendicularly out of page as shown. Find out the magnetic field at (x,y)= (2a,0).field at (x,y) (2a,0).

Solution:Draw a “Ampere Circle” that centers yDraw a “Ampere Circle” that centers

at (0,0) and passes (2a,0).Apply Ampere’s Law

y

(0 0) (2a 0)xIB

IrBd

enclosed

enclosed

μ

μπ2

0

0

⇒

==•∫ sBr

Take r=2a, Ienclosed=I

(0,0) (2a,0)r

B enclosed

πμ

20=⇒

Idirection? up

aIB

πμ4

0=

Field Test -- Exercise 9 (Part B) Now add another infinite straight current I at (x,y)= (a,0) that also runs perpendicularly out of page. Again, find out the magnetic field at (x,y)= (2a,0).magnetic field at (x,y) (2a,0).

Solution:Draw a “Ampere Circle” that centers yDraw a “Ampere Circle” that centers

at (0,0) and passes (2a,0).Apply Ampere’s Law

y

(0 0) (2a 0)xIB

IrBd

enclosed

enclosed

μ

μπ2

0

0

⇒

==•∫ sBr

(a 0)

Take r=2a, Ienclosed=I + I =2I

(0,0) (2a,0)r

B enclosed

πμ

20=⇒ (a,0)

direction up aIB

πμ2

0=

Field Test -- Exercise 9 (Part B) Now add another infinite straight current I at (x,y)= (a,0) that also runs perpendicularly out of page. Again, find out the magnetic field at (x,y)= (2a,0).magnetic field at (x,y) (2a,0).

Correct solution:The current at (0 0) contributes yThe current at (0,0) contributes

(dir. up) at (2a,0).

y

aIB

πμ4

01 =

By same argument, the current at (a,0)contributes at (2a,0) (0 0) (2a 0)

x(a 0)

IB μ 0co bu es ( ,0)

Total B

(0,0) (2a,0)(a,0)a

Bπ2

02 =

Iμ3(up)a

IBBBπμ

43 0

21 =+=