Physics 2102 Physics 2102

Exam 2: Review SessionExam 2: Review Session

Chapters 24.9-28.8 / HW04-06

Physics 2102

Jonathan Dowling

Some links on exam stress:http://appl003.lsu.edu/slas/cas.nsf/$Content/Stress+Management+Tip+1http://wso.williams.edu/orgs/peerh/stress/exams.htmlhttp://www.thecalmzone.net/Home/ExamStress.phphttp://www.staithes.demon.co.uk/exams.html

Exam 2Exam 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 ResistanceCurrent 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.

Potential V of Continuous Charge Distributions

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dV =kdqr

V = dV∫

dV =kdqr

V = dV∫Straight Line Charge: dq=dx=Q/L

Curved Line Charge: dq=ds=Q/2R

Surface Charge: dq=dA=Q/R2

dA=2R’dR’

Straight Line Charge: dq=dx=Q/L

Curved Line Charge: dq=ds=Q/2R

Surface Charge: dq=dA=Q/R2

dA=2R’dR’

r = ′R 2 + z2r = ′R 2 + z2

Potential V of Continuous Charge Distributions

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Straight Line Charge: dq=dx=Q/LStraight Line Charge: dq=dx=Q/L

Curved Line Charge: dq=ds=Q/2RCurved Line Charge: dq=ds=Q/2R

Potential V of Continuous Charge Distributions

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Surface Charge: dq=dA=Q/R2

dA=R’dR’d

Straight Line Charge: dq=dx=bx is given to you.

Straight Line Charge: dq=dx=bx is given to you.

CapacitorsCapacitors

E = /0 = q/A0

E = V dq = C V

C = 0A/d

C = 0A/d

C=0ab/(b-a)

Connected to Battery: V=ConstantDisconnected: Q=Constant

Connected to Battery: V=ConstantDisconnected: Q=Constant

Current and ResistanceCurrent and Resistancei = dq/dt

V = i RE = J = 0(1+(TT0))

R = L/AJunction rule

DC CircuitsDC Circuits

Single loop Multiloop

V = iRP = iV

Loop rule

Resistors and CapacitorsResistors and Capacitors

Resistors Capacitors

Key formula: V=iR Q=CV

In series: same current same charge

Req=∑Rj 1/Ceq= ∑1/Cj

In parallel: same voltage same voltage

1/Req= ∑1/Rj Ceq=∑Cj

Capacitors and ResistorsCapacitors and Resistorsin Series and in Parallelin 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?

RC CircuitsRC Circuits Time constant: RC

( )/

/0

Charging: ( ) 1

Discharging: ( )

t RC

t RC

q t CE e

q t q e

−

−

= −

=

i(t)=dq/dt

Capacitors: Checkpoints, Capacitors: Checkpoints, QuestionsQuestions

Problem 25-21

When switch S is thrown to the left, the plates of capacitor 1 acquire a potential V0. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges q1, q2, and q3 on the capacitors?

Series

SeriesSeri-Q

Current and Resistance: Checkpoints, QuestionsCurrent and Resistance: Checkpoints, Questions

Problem 26-56

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 m. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?

Circuits: Checkpoints, QuestionsCircuits: Checkpoints, Questions

Problem: 27.P.018. [406649]Figure 27-33 shows five 5.00 resistors. (Hint: For each pair of points, imagine that a battery is connected across the pair.)

Fig. 27-33(a) Find the equivalent resistance between points F and H.

(b) Find the equivalent resistance between points F and G.

Problem: 27.P.046. [406629]

In an RC series circuit, E = 17.0 V, R = 1.50 M, 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?

Magnetic Forces and TorquesMagnetic Forces and Torques

F qv B q E= × +r r rr

BLdiFdrrr

×=

Brrr

×=μτ

L

vF qB

mvr =

C

C

Top viewSide view

Consider the rectangular loop in fig. with sides of lengths and and that carries

ˆa current . The loop is placed in a magnetic field so that the normal to th

a a b

i n

Magnetic Torque on a Current Loop

1 3

2 4

e loop

forms an angle with . The magnitude of the magnetic force on sides 1 and 3 is

sin 90 . The magnetic force on sides 2 and 4 is

sin(90 ) cos . These forces cancel

B

F F iaB iaB

F F ibB ibB

θ

θ θ= = °== = − =

r

net

2 4

1 3

in pairs and thus 0.

The torque about the loop center of and is zero because both forces pass

through point . The moment arm for and is equal to ( /2)sin . The two

torques tend to

F

C F F

C F F b θ

=

1 3

rotate the loop in the same (clockwise) direction and thus add up.

The net torque + =( /2)sin ( /2)sin sin sin .iabB iabB iabB iABτ τ τ θ θ θ θ= + = =

net siniABτ θ=

net 0F =

(28-13)

U Bμ= U Bμ=−

The torque of a coil that has loops exerted

by a uniform magnetic field and carries a

current is given by the equation .

We define a new vector associated with the

N

B

i NiABτμ

=

Magnetic Dipole Moment

r coil,

which is known as the magnetic dipole moment of

the coil.

The magnitude of the magnetic dipole moment is

Its direction is perpendicular to the plane of the coil.

The sense of is defined by the right-hand rule. We curl the fingers of the right an

i

.

h d

NiAμ

μ

=

r

n the direction of the current. The thumb gives us the sense. The torque can be

expressed in the form sin where is the angle between and .

In vector form:

The potential energy

.

B

B

Bτ μ θ θ μ

τ μ=

=

×

r

rr r

r

of the coil is:

has a minimum value of for 0 (position of equilibrium).

has a maximum value of for 180 (position of equilib

cos

rium).

For both posit

.

U B

U

B B

B

U

μ θ

μ θ

μ θ μ

−

= ⋅

=

°

− =−

=

stable

unstable

Note:

rr

ions the net torque is 0.τ =

Bτ μ= ×rr r

(28-14)

U Bμ=− ⋅rr

Ch 28: Checkpoints and QuestionsCh 28: Checkpoints and Questions

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

B into blackboard.

v

F

€

FB = FC

→ qv B =mv 2

r

€

FB = FC

→ qv B =mv 2

r

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

€

Fcentrifugalout = ma = mrω2 = m

v 2

r

€

Fcentrifugalout = ma = mrω2 = m

v 2

r

€

Fmagneticin = qvB

€

Fmagneticin = qvB

€

Solve : r =mv

qB

€

Solve : r =mv

qB

.

.electron

rB

vr

Fr

C

r

€

r =mv

qB

€

r =mv

qB

€

ω =v

r=qB

m

€

ω =v

r=qB

m

€

T ≡2πr

v=

2πmv

qBv=

2πm

qB

€

T ≡2πr

v=

2πmv

qBv=

2πm

qB

€

f ≡1

T=qB

2πm

€

f ≡1

T=qB

2πm

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

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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