Lecture 31: MON 30 MAR Review Session : Midterm 3

Physics 2113 Jonathan Dowling

EXAM 03: 8PM MON 30 MAR in Cox Auditorium

The exam will cover: Ch.26 through Ch.29 The exam will be based on: HW07 – HW10.

The formula sheet for the exam can be found here: http://www.phys.lsu.edu/classes/fall2014/phys2113/downloads/FormulaSheetTest3.pdf

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

Current and Resistance i = dq/dt

V = i R E = J ρ

r = ρ0(1+a(T-T0))

R = ρL/A Junction rule

Current and Resistance

Current and Resistance: Checkpoints, Questions

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?

Circuits i = dq/dt

Junction rule

DC Circuits

Single loop Multiloop

V = iR P = iV

Loop rule

Resistors vs 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

Resistors in Series and in Parallel

•  What’s the equivalent resistance? •  What’s the current in each resistor? •  What’s the potential across each resistor? •  What’s the current delivered by the battery? •  What’s the power dissipated by each resisitor?

Series and Parallel

Series and Parallel

Series and Parallel

Circuits: 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. Slide Rules:

You may bend the wires but not break them. You may slide any circuit element along a wire so long as you don’t slide it past a three (or more) point junction or another circuit element.

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?

Magnetic Forces and Torques

F qv B q E= × +! ! !!

BLdiFd!!!

×=

B!!! ×= µτ

L

v F qB

mvr =

C

C

Top view Side view

Consider the rectangular loop in fig. with sides of lengths and and that carriesˆa current . The loop is placed in a magnetic field so that the normal to th

a a bi 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 issin 90 . The magnetic force on sides 2 and 4 is sin(90 ) cos . These forces cancel

BF F iaB iaBF F ibB ibB

θ

θ θ= = ° == = − =

!

net

2 4

1 3

in pairs and thus 0.The torque about the loop center of and is zero because both forces passthrough point . The moment arm for and is equal to ( / 2)sin . The twotorques tend to

FC 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 exertedby a uniform magnetic field and carries a current is given by the equation .We define a new vector associated with the

NB

i NiABτµ

=

Magnetic Dipole Moment

! 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 ani

.

h d

NiAµ

µ

=

!

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

.

BBBτ µ θ θ µ

τ µ=

=

×

!

!! !

!

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 BU

B B

B

Uµ θ

µ θ

µ θ µ−

= ⋅=

°

− = −

=stableunstable

Note :

!!

ions the net torque is 0.τ =

Bτ µ= ×!! !

(28-14)

U Bµ= − ⋅!!

Ch 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

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

r

�

Fcentrifugalout = ma = mrω 2 = m v 2

r

�

Fmagneticin = qvB

�

Solve : r = mvqB

.

. electron

!B

v!

F!

C r

�

r = mvqB

�

ω = vr

= qBm

�

T ≡ 2πrv

= 2πmvqBv

= 2πmqB

�

f ≡ 1T

= qB2πm

Radius of Circlcular Orbit

Angular Frequency: Independent of v

Period of Orbit: 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?

!µ

!µ Highest Torque:

θ = ±90° sinθ = ±1

Lowest Torque: θ= 0° & 180° sinθ = 0

B θ = 180° –cosθ = +1

θ = 0° –cosθ = –1

!τ = µBsinθ

= −µBcosθ

!F = iLBsinθ

Right Hand Rule: Given Current i Find Magnetic Field B 

Checkpoints/Questions Magnetic field?

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

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

A

l Px

y

z

C

B D +z is out of page

BB BA

BD BCBB,C BD

BABD

BA BB,C

BB,C BA

BDBA

BD BB,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).

A

l Px

y

z

C

B D +z is out of page

BB BA

BD BCBB,C BD

BABD

BA BB,C

BB,C BA

BDBA

BD BB,C

Right Hand Rule: Given Current i Find Magnetic Field B 

A length of wire is formed into a closed circuit with radii a and b, as shown in the Figure, and carries a current i.

(a) What are the magnitude and direction of B at point P? (b) Find the magnetic dipole moment of the circuit.

RiBπφµ

40=

µ=NiA

Right Hand Rule & Biot-Savart: Given i Find B 

!B = ⊗