Lecture 33: WED 11 NOV Review Session B: Midterm 3

Physics 2113 Jonathan Dowling

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: 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/

The Biot-Savart Law For B-Fields

•  Quantitative rule for computing the magnetic field from any electric current

•  Choose a differential element of wire of length dL and carrying a current i

•  The field dB from this element at a point located by the vector r is given by the Biot-Savart Law

dB! "!

= µ04π

idL! "!

× "rr3

= µ04π

idL! "!

× r̂r2

µ0 =4π×10–7 T•m/A (permeability constant of free space)

Jean-Baptiste Biot (1774-1862)

Felix Savart (1791-1841) dL

! "!

!r

i

dB! "!

⊗

( )∫∞

∞− += 2/322

0

4 RsRdsi

πµ

( )∫∞

+=

02/322

0

2 RsRdsi

πµ

( )∞

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

0

2/12220

2 RsRsiR

πµ

Ri

πµ20=

sinθ = cos(θ −π / 2) = R / r 2/122 )( Rsr +=

∫∞

∞−

= 30 )sin(4 r

rdsiB θπµ

Field of a Straight Wire

d!B = µ0

4πid!s × !rr3 3

0 )sin(4 r

ridsdB θπµ=

π −θ

θ −π / 2

Biot-Savart Law •  A circular arc of wire of radius R

carries a current i. •  What is the magnetic field at the

center of the loop?

d!B = µ0

4πids"!"

× !rr3

20

30

44 RiRd

RidsRdB φ

πµ

πµ ==

0 0

4 4iidB

R Rµ µφπ π

Φ= =∫

Direction of B?? Not another right hand rule?! TWO right hand rules!: If your thumb points along the CURRENT, your fingers will point in the same direction as the FIELD. If you curl our fingers around direction of CURRENT, your thumb points along FIELD!

i

Example, Magnetic field at the center of a circular arc of a circle pg 768:

ICPP: What is the direction of the B field at point C?

(a) Into the board ⊗?(b) Out of the board ⊙?(c) To the right → ?(d) To the left ← ?

✔

⊗

Example, Magnetic field at the center of a circular arc of a circle.:

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 = ⊗

B1 =µ0I12πd

Magnetic field due to wire 1 where the wire 2 is,

1221 BILF =

d

I2 I1 L

= µ0LI1 I22πd

Force on wire 2 due to this field, F

Forces Between Wires

eHarmony’s Rule for Currents: Same Currents – Attract! Opposite Currents – Repel!

29.3: Force Between Two Parallel Wires:

⊙a

F21 =µ0Li1i2

2πd= Force on 2 due to 1.

Like currents attract & Opposites Repel.1/r ⇒ double the distance halve the force.

⊙b

⊗c F

!"ab

F!"anet

⊙a

⊙b

⊗c

F!"bnet

F!"bc

F!"ba

⊙a

⊙b

⊗c

F!"cb

F!"cnet

F!"ca

Fbnet > Fc

net > Fanet

F!"ac

29.3.3. The drawing represents a device called Roget’s Spiral. A coil of wire hangs vertically and its windings are parallel to one another. One end of the coil is connected by a wire to a terminal of a battery. The other end of the coil is slightly submerged below the surface of a cup of mercury. Mercury is a liquid metal at room temperature. The bottom of the cup is also metallic and connected by a wire to a switch. A wire from the switch to the battery completes the circuit. What is the behavior of this circuit after the switch is closed?

a) When current flows in the circuit, the coils of the wire move apart and the wire is extended further into the mercury. b) Nothing happens to the coil because there will not be a current in this circuit. c) A current passes through the circuit until all of the mercury is boiled away. d) When current flows in the circuit, the coils of the wire move together, causing the circuit to break at the surface of the mercury. The coil then extends and the process begins again when the circuit is once again complete.

29.3: Force Between Two Parallel Wires, Rail Gun:

The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current traversing the loop.

i1 i2

i3

ds

i4

)( 3210loop

iiisdB −+=⋅∫ µ!!

Thumb rule for sign; ignore i4

If you have a lot of symmetry, knowing the circulation of B allows you to know B.

Ampere’s law: Closed Loops

!B ⋅d!s

LOOP"∫ = µ0ienclosed

The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current piercing the loop.

�

! B ⋅d! s

loop∫ = µ0 (i1 − i2)

Thumb rule for sign; ignore i3

If you have a lot of symmetry, knowing the circulation of B allows you to know B.

Ampere’s law: Closed Loops

!B ⋅d!s

LOOP"∫ = µ0ienclosed

e n c C a l c u l a t i o n o f . W e c u r l t h e f i n g e r s o f t h e r i g h t h a n d i n t h e d i r e c t i o n i n w h i c h t h e A m p e r i a n l o o p w a s t r a v e r s e d . W e n o t e t h e

i

d i r e c t i o n o f t h e t h u m b .

All currents inside the loop to the thumb are counted as .All currents inside the loop to the thumb are counted as .All currents outside the loop are not count

paantiparalrallel positive

lel negative

enc 1 2

ed. In this example : .i i i= −

(a) − i + i + i = i

(b) − i + 0 + i = 0

(c) − i + 0 + 0 = −i(d) 0 + i + i = 2i

d > a = c > b = 0

Ampere’s Law: Example 2

•  Infinitely long cylindrical wire of finite radius R carries a total current i with uniform current density

•  Compute the magnetic field at a distance r from cylinder axis for: r < a (inside the wire) r > a (outside the wire)

i Current out

of page, circular field

lines

∫ =⋅C

isdB 0µ!!

Ampere’s Law: Example 2 (cont)

∫ =⋅C

isdB 0µ!!

Current out of page, field

tangent to the closed

amperian loop enclosedirB 0)2( µπ =

2

22

22 )(

Rrir

Ri

rJienclosed === ππ

πr

iB enclosed

πµ20=

20

2 RirB

πµ= For r < R For r>R, ienc=i, so

B=µ0i/2πR = LONG WIRE!

Need Current Density J!

R

0

2iR

µπ

r

B

O

B-Field In/Out Wire: J is Constant

r < RB ∝ r

r > RB ∝1 / r

r < R

B =µ0ir2πR2

r > R

B =µ0i2πr

Outside Long Wire!

!B ⋅d!s

LOOP"∫ = µ0ienclosed

!∫ a∝ 4 = 4

!∫ b∝ 4 − 9 = 5

!∫ c∝ 4 − 9 + 5 = 0

!∫ d∝ 4 − 9 + 5 − 3 = 3

+4⊙

−9⊗ +5⊙

3⊗

!∫ b> !∫ a

> !∫ d> !∫ c

= 0

B = µ0ienclosed2πr

Ba ∝ 4 /1= 4 = 16 / 4Bb ∝ 4 − 9 / 2 = 5 / 2 = 10 / 4

Bc ∝ 4 − 9 + 5 / 3= 0

Bd ∝ 4 − 9 + 5 − 3 / 4 = 3 / 4

+4⊙

−9⊗ +5⊙

3⊗

Ba > Bb > Bd > Bc = 0

B

1/r Law too

ra = 1m

29.4.1. A copper cylinder has an outer radius 2R and an inner radius of R and carries a current i. Which one of the following statements concerning the magnetic field in the hollow region of the cylinder is true?

a) The magnetic field within the hollow region may be represented as concentric

circles with the direction of the field being the same as that outside the cylinder. b) The magnetic field within the hollow region may be represented as concentric

circles with the direction of the field being the opposite as that outside the cylinder.

c) The magnetic field within the hollow region is parallel to the axis of the cylinder

and is directed in the same direction as the current. d) The magnetic field within the hollow region is parallel to the axis of the cylinder

and is directed in the opposite direction as the current. e) The magnetic field within the hollow region is equal to zero tesla.

29.4.3. The drawing shows two long, straight wires that are parallel to each other and carry a current of magnitude i toward you. The wires are separated by a distance d; and the centers of the wires are a distance d from the y axis. Which one of the following expressions correctly gives the magnitude of the total magnetic field at the origin of the x, y coordinate system?

a) b) c) d) e) zero tesla

0

2id

µ

0

2id

µ

0

2id

µπ

0id

µπ

29.5: Solenoids:

Fig. 29-19 Application of Ampere’s law to a section of a long ideal solenoid carrying a current i. The Amperian loop is the rectangle abcda.

Here n be the number of turns per unit length of the solenoid

29.5.5. A solenoid carries current I as shown in the figure. If the observer could “see” the magnetic field inside the solenoid, how would it appear?

29.5.1. The drawing shows a rectangular current carrying wire loop that has one side passing through the center of a solenoid. Which one of the following statements describes the force, if any, that acts on the rectangular loop when a current is passing through the solenoid.

a) The magnetic force causes the loop to move upward. b) The magnetic force causes the loop to move downward. c) The magnetic force causes the loop to move to the right. d) The magnetic force causes the loop to move to the left. e) The loop is not affected by the current passing through the solenoid or the

magnetic field resulting from it.

→i

Magnetic Field of a Magnetic Dipole

30

2/3220

2)(2)(

zzRzB µ

πµµ

πµ !!!

≈+

=

All loops in the figure have radius r or 2r. Which of these arrangements produce the largest magnetic field at the point indicated?

A circular loop or a coil currying electrical current is a magnetic dipole, with magnetic dipole moment of magnitude µ=NiA. Since the coil curries a current, it produces a magnetic field, that can be calculated using Biot-Savart’s law:

29.6: A Current Carrying Coil as a Magnetic Dipole:

For small z << R

B(z) = µ0iR2R3 1+ z2 / R2( )3/2 ≅ µ0i

2R2

1 / R2 Law! Double the R one fourth the field.

29.6: A Current Carrying Coil as a Magnetic Dipole:

It's a 1/r3 law!!!

Lenz’s Law

Induction and Inductance •  Faraday’s law: or

•  Inductance: L=NΦ/i –  For a solenoid: L=µ0n2Al=µ0N2A/l

•  Inductor EMF: EL= -L di/dt •  RL circuits: i(t)=(E/R)(1–e–tR/L) or i(t)=i0e–tR/L

•  RL Time Constant: τ = L/R Units: [s]

•  Magnetic energy: U=Li2/2 Units: [J] •  Magnetic energy density: u=B2/2m0 Units: [J/

m3]

E = −

dΦB

dt dtdsdE B

C

Φ−=⋅∫!!

i

Summary Two versions of Faradays’ law:

–  A time varying magnetic flux produces an EMF:

�

E = − dΦB

dt

dtdsdE B

C

Φ−=⋅∫!!

– A time varying magnetic flux produces an electric field:

Changing B-Flux Induces EMF

iup t( ) = ER

1− e− t /τ L( )UB t( ) = 1

2Li2

= 12L E 2

R2 1− e− t /τ L( )2

To find ELwalk the loop :+E +VR +EL = 0EL = −E −VR = −E − −iR( )

= −E + ER

1− e− t /τ L( )⎡⎣⎢

⎤⎦⎥R

EL = −E e− t /τ L

Flux Up Flux Down

idn t( ) = ERe− tR/L

UB t( ) = 12Li2

= 12L E 2

R2 e−2tR/L

Walk the loop!EL = −VR = −iREL = −E e− t /τ L

RL Circuits

τ L = L / R

RL Circuits

E/2

t=?

Make or Break? At t=0 all L’s make breaks so throw out all loops with a break and solve circuit. At t=∞ all L’s make solid wires so replace L’s with wire and solve circuit.

Checkpoints/Questions Magnitude of induced emf/current? Magnitude/direction of induced current?

Magnitude/direction of magnetic field inducing current?

Given |∫E·ds| , direction of magnetic field?

Current inducing EL?

Current through the battery? Time for current to rise 50% of max value?

Given B, dB/dt, magnitude of electric field?

Largest current?

Largest L?

R,L or 2R,L or R, 2L or 2R,2L?

EEMF = dΦB

dt= A dB

dtEMF is proportional to the slope dB/dt

Eb > Ed = Ee > Ea = Ec = 0

Ia = Ib > Ic = 0

30.4.1. Consider the situation shown. A triangular, aluminum loop is slowly moving to the right. Eventually, it will enter and pass through the uniform magnetic field region represented by the tails of arrows directed away from you. Initially, there is no current in the loop. When the loop is entering the magnetic field, what will be the direction of any induced current present in the loop?

a) clockwise b) counterclockwise c) No current is induced.

30.4.3. A rigid, circular metal loop begins at rest in a uniform magnetic field directed away from you as shown. The loop is then pulled through the field toward the right, but does not exit the field. What is the direction of any induced current within the loop?

a) clockwise b) counterclockwise c) No current is induced.

ICPP •  3 loops are shown. •  B = 0 everywhere except in

the circular region I where B is uniform, pointing out of the page and is increasing at a steady rate.

•  Rank the 3 loops in order of increasing induced EMF. –  (a) III < II < I ?

–  (b) III < II = I ?

–  (c) III = II = I ?

• III encloses no flux so EMF=0 • I and II enclose same flux so EMF same. •  Are Currents in Loops I & II Clockwise or Counterclockwise?

I II

III

B

d = c > b = a

E = − dΦdt

= −L didt

If i is constant (in time) then E = 0

a?b?

d?e?c?

f ?

The RL circuit •  Set up a single loop series circuit

with a battery, a resistor, a solenoid and a switch.

•  Describe what happens when the switch is closed.

•  Key processes to understand: –  What happens JUST AFTER

the switch is closed? –  What happens a LONG TIME

after switch has been closed? –  What happens in between?

Key insights: •  You cannot change the CURRENT in

an inductor instantaneously! •  If you wait long enough, the

current in an RL circuit stops changing!

At t = 0, a capacitor acts like a solid wire and inductor acts like break in the wire. At t = ∞ a capacitor acts like a break in the wire and inductor acts like a solid wire.

ICPP Immediately after the switch is closed, what is the potential difference across the inductor? (a) 0 V (b) 9 V (c) 0.9 V

•  Immediately after the switch, current in circuit = 0. •  So, potential difference across the resistor = 0! •  So, the potential difference across the inductor = E = 9 V!

10 Ω

10 H 9 V

At t = 0 when you walk the loopsince i(0) = 0 it's like VR = iR is not even there!E −VL = 0VL = E = 9.0V

Fluxing Up The Inductor

•  How does the current in the circuit change with time?

−iR +E − L di

dt= 0

i = E

R1− e− t /τ( )

Time constant of RL circuit: τLR = L/R t

E/R

i(t) Fast = Small τ

Slow= Large τ

i

Fluxing Down an Inductor

0=+dtdiLiR

i = E

Re−Rt /L = E

Re− t /τ t

E/R Exponential defluxing i(t)

i

The switch is at a for a long time, until the inductor is charged. Then, the switch is closed to b. What is the current in the circuit? Loop rule around the new circuit walking counter clockwise:

τ RL = L / R

When the switch is closed, the inductor begins to get fluxed up, and the current is i=(E/R)(1-e–t/τ). When the switch is opened, the inductors begins to deflux.

The current in this case is then i= (E/R) e–t/τ

Example, RL circuit, immediately after switching and after a long time: SP30.05