phy hw
fkmyhomework
Week3.zipWeek 3/.DS_Store
__MACOSX/Week 3/._.DS_Store
Week 3/Week 3/Exam 2.pdf
1. An electromagnetic wave is traveling west with electric field vector oscillating vertically up and down. Its magnetic field vector is oscillating
a. north-south. b. east-west. c. vertically up and down.
2. Which of these types of EM waves has the longest wavelength in vacuum?
a. x-rays. b. infrared. c. red light. d. blue light. e. ultraviolet. f. AM radio. g. all tie.
3. T F Garage door openers use lower frequency EM waves than wireless internet technology.
4. T F Microwave ovens use EM waves of about the same frequency as wireless internet technology.
5. T F The main absorber of ultraviolet light in Earth’s atmosphere is methane (CH4).
6. A TV dish antenna receives EM waves of frequency 2.55 GHz from a satellite 36000 km away.
b) What is the distance between successive wave crests of the electric field vector of the waves?
a) How much time in milliseconds does the signal require to travel from the satellite to the dish antenna?
7. A light source emits EM waves of 4.00 W total power uniformly in all directions. Assume the waves have a single frequency (monochromatic).
a) What is the intensity of the waves at 2.00 m from the light source?
b) What is the electric field amplitude E0 produced at 2.00 m from the light source?
1
19. Two narrow slits are separated by 0.250 mm and illuminated with green light of wavelength 555 nm. The light passes through the slits and shines on a screen 2.00 m behind the slits. 8. If the zeroth order bright fringe falls on the screen at angle θ = 0 radians (straight in front of the slits), at what angle (in radians) does the first order bright fringe appear?
b) How far apart are neighboring bright fringes on the screen?
Prefixes
a=10−18, f=10−15, p=10−12, n=10−9, µ = 10−6, m=10−3, c=10−2, k=103, M=106, G=109, T=1012, P=1015
Physical Constants
k = 1/4π�0 = 8.988 GNm 2/C2 (Coulomb’s Law) �0 = 1/4πk = 8.854 pF/m (permittivity of space)
e = 1.602 × 10−19 C (proton charge) µ0 = 4π × 10−7 T·m/A (permeability of space) me = 9.11 × 10−31 kg (electron mass) mp = 1.67 × 10−27 kg (proton mass) c = 3.00 × 108 m/s (speed of light) c = 2.99792458 × 108 m/s (exact value in vacuum)
Units
NA = 6.02 × 1023/mole (Avogadro’s #) 1 u = 1 g/NA = 1.6605 × 10−27 kg (mass unit) 1.0 eV = 1.602 × 10−19 J (electron-volt) 1 V = 1 J/C = 1 volt = 1 joule/coulomb 1 F = 1 C/V = 1 farad = 1 C2/J 1 H = 1 V·s/A = 1 henry = 1 J/A2 1 A = 1 C/s = 1 ampere = 1 coulomb/second 1 Ω = 1 V/A = 1 ohm = 1 J·s/C2 1 T = 1 N/A·m = 1 tesla = 1 newton/ampere·meter 1 G = 10−4 T = 1 gauss = 10−4 tesla
Electromagnetic waves:
|~E|/|~B| = c = 1/ √ �0µ0, (fields) fλ = c (wave equation)
Energy density, intensity, power:
u = �0E 2 = B
2
µ0 (instantaneous energy density) u = 1
2 �0E
2 0 =
B20 2µ0
(average energy density)
I = uc = 1 2 �0E
2 0c (EM waves intensity) I = P/A = P/(4πr
2) (intensity definition)
Reflection, Mirrors:
θr = θi (angle of incidence = angle of reflection) f = r/2 (focal length of spherical mirror) 1 do
+ 1 di
= 1 f
(mirror equation) m = −di/do = hi/ho (linear magnification) di > 0 ⇒ real, light side. di < 0 ⇒ virtual, dark side. m > 0 ⇒ upright. m < 0 ⇒ inverted. |m| > 1 ⇒ magnified. |m| < 1 ⇒ diminished.
Refraction, Lenses:
n = c/v (index of refraction) n1 sin θ1 = n2 sin θ2 (Snell’s Law) 1 do
+ 1 di
= 1 f
(lens equation) m = −di/do = hi/ho (linear magnification) di > 0 ⇒ real image, light (opp.) side. di < 0 ⇒ virtual image, dark (same) side. m > 0 ⇒ upright. m < 0 ⇒ inverted. |m| > 1 ⇒ magnified. |m| < 1 ⇒ diminished.
Eq.-1
λn = λvacuum/n (wavelength in a medium) ∆x = d sin θ (path difference in double slits)
d sin θ = mλ (double slits bright fringes) d sin θ = (m + 1/2)λ (double slits dark fringes)
Diffraction:
D sin θ = mλ (single slit minima) y = L tan θ (position on a screen)
d sin θ = mλ (diffraction grating maxima) d = 1/(lines per meter).
Thin film interference:
∆x = path 1© - path 2© (path difference) reflect from higher n ⇒ extra λ/2 path change. ∆x = mλ (constructive interference) ∆x = (m + 1/2)λ (destructive interference)
Polarization:
I = I0 cos 2 θ (transmission thru polarizer) I = 1
2 I0 (transmission of unpolarized light)
f/D = f-number, or lens aperture film exposure = exposure time / f-number. 1 do
+ 1 di
= 1 f
(lens equation) m = −di/do = hi/ho (image size and magnification)
Lens power
P = 1/f (power in diopters).
Vision correction
Far point FP = ∞. (good vision) Near point = NP ≤ 25 cm. (good vision) Nearsighted. Use lens to get FP=∞. Farsighted. Use lens to get NP=25 cm.
Simple magnifier
θ = h0 NP
(angular size viewed at NP.) θ′ = h0 d0
(angular size viewed at any d0.)
M = θ ′
θ = NP
do (ang. Mag. viewed at any d0.) M =
θ′
θ = NP
f (ang. Mag. viewed at d0 = f.)
Telescopes
M = θ ′
θ =
fobj feye
(angular magnification)
Eq.-2
Week 3/Week 3/Exam 1.docx
EXAM 1
Q1. A magnetic field points perpendicularly into a 2.0 cm × 4.0 cm rectangular coil with 120 turns and 0.25 Ω total resistance. At a certain instant, the magnetic field strength is 0.22 T and it is increasing at a rate of 2.5 T/s.
a) The induced current in the coil is
a. clockwise. b. counter clockwise.
b) Calculate the instantaneous induced current in the coil.
Q2. Two long straight wires carry currents perpendicular to the page as shown. Determine the x and y components of the net magnetic field produced at point P.
Q3. A 40.0-meter long straight wire carries a 75.0 A current towards the west. Earth’s 0.45-gauss magnetic field there points north but 55.0◦ below horizontal.
a) In what direction is the magnetic force F⃗ on the wire? Draw and label F⃗ on the diagram or use ⊙ to show that F⃗ is out-of-the-page or ⊗ to show that F⃗ is into-the-page.
b) Calculate the magnitude of the magnetic force on the wire.
Q4. A fluorine ion (q = +4e, m = 3.15 × 10−26 kg) that was accelerated through a potential difference of 48.0 kV moves in circular motion in a uniform 0.850-tesla magnetic field, see diagram.
a) T rue False The diagram shows the correct direction of its motion.
b) Calculate the speed of the fluorine ion.
c) Calculate the radius of its cyclotron orbit.
__MACOSX/Week 3/Week 3/._Exam 1.docx
Week 3/Week 3/6.pdf
1
Electromagnetic Induction
Chapter 21
Faraday’s Law
Part 1
Faraday’s observation
Faraday: If the magnetic field changes, or if the magnet and coil are in relative motion, there will be an induced emf (and therefore current) in the coil.
Electric currents produce magnetic fields. 19th century puzzle: Can magnetic fields produce currents? A static magnet will produce no current in a stationary coil.
Magnetic Flux For a “loop” of wire (not necessarily circular) with area A, in an external magnetic field B, the magnetic flux through the loop is:
SI units of Magnetic Flux: 1 T·m2 = 1 weber = 1 Wb
A = area of loop θ = angle between B and the normal to the loop
ϑcosBA=Φ
Induced emf (Voltage) from changing Magnetic Flux
Faraday: If the magnetic field changes, or if the magnet and coil are in relative motion, there will be an induced emf (and therefore current) in the coil.
Key Concept: The magnetic flux through the coil must change, this will induce an emf e in the coil, which produces a current I = emf/R in the coil.
Such a current is said to be induced by the time varying magnet flux that “links” the coil.
Problem A magnetic field is oriented at an angle of 32º to the normal of a rectangular area 5.5 cm by 7.2 cm. If the magnetic flux through this surface has a magnitude of 4.8 × 10-5 T·m2, what is the strength of the magnetic field?
5
cos or cos
4.8 10 14.3 mT
cos 0.055 0.072 cos 32
BA B A
B A
θ θ
θ
−
Φ Φ = =
Φ × = = =
× ×
2
Faraday’s Law of Induction
Faraday’s Law: The instantaneous EMF (voltage) induced in a circuit (w/ N loops) equals the rate of change of magnetic flux through the circuit:
if
if tt
N t
N − Φ−Φ
−= Δ ΔΦ
−=ε
The minus sign just indicates the direction of the induced emf. To calculate the magnitude, we will use:
if
if tt
N t
N − Φ−Φ
= Δ ΔΦ
=ε
Faraday’s Law of Induction
if
iiifff
if
if
tt ABAB
tt − −
−= − Φ−Φ
−= θθ
ε coscos
Induction by relative motion
if
iiifff
if
if
tt ABAB
tt − −
−= − Φ−Φ
−= θθ
ε coscos
if
if
tt BB
A − −
−= θε cos
Induction by Rotational Motion
As a coil rotates in a constant magnetic field (uniform or not) the flux through the loop changes, inducing an emf in the coil.
if
iiifff
if
if
tt ABAB
tt − −
−= − Φ−Φ
−= θθ
ε coscos
if
if
tt BA
− −
−= θθ
ε coscos
Good to discuss The graph shows the magnitude B of a uniform magnetic field that is perpendicular to the plane of a conducting loop. Rank the five regions indicated on the graph according to the magnitude of the emf induced in the loop, from least to greatest.
Problem This is a plot of the magnetic flux through a coil as a function of time. At what times shown in this plot does (a) the magnetic flux and (b) the induced emf have the greatest magnitude?
3
Problem A 0.25 T magnetic field is perpendicular to a circular loop of wire with 50 turns and a radius 15 cm. The magnetic field is reduced to zero in 0.12 s. What is the magnitude of the induced emf?
( )( )2 and
0 0.25 0.15 50 50 7.36 volts
0.12 0.12 f i
emf N emf N t t
emf π
ΔΦ ΔΦ = − =
Δ Δ
− × ×Φ − Φ = = =
Lenz’s Law
Part 2
Lenz’s Law
Lenz’s Law: An induced current always flows in a direction that opposes the change that caused it.
In this example the magnetic field in the downward direction through the loop is increasing. So a current is generated in the loop which produces an upward magnetic field inside the loop to oppose the change.
Magnet moving down toward loop
N
S
Induced current
Induced B field
Induction by Relative Motion
• When a permanent magnet moves relative to a coil, the magnetic flux through the coil changes (WHY?), inducing an emf in the coil.
• In a) the magnitude of the flux is increasing
• In c) the flux is decreasing in magnitude.
• In a) and c) the induced current is in the opposite direction (Lenz’s law).
v
v
Lenz’s Law Good to discuss A square loop of wire lies in the plane of the page. A decreasing magnetic field is directed into the page. The induced current in the loop is: A) counterclockwise B) clockwise C) zero D) depends upon whether or not B is decreasing at a constant rate E) clockwise in two of the loop sides and counterclockwise in the other two
4
Good to discuss A long, straight wire carries a steady current I. A rectangular conducting loop lies in the same plane as the wire, with two sides parallel to the wire and two sides perpendicular. Suppose the loop is pushed toward the wire as shown. Given the direction of I, the induced current in the loop is 1. clockwise. 2. counterclockwise. 3. need more information
Problem The figure shows a circuit containing a resistor and an uncharged capacitor. Pointing into the plane of the circuit is a uniform magnetic field B. If the magnetic field increases in magnitude with time, which plate of the capacitor (top or bottom) becomes positively charged? The magnetic flux through the circuit is changing so that there are more magnetic field lines into the paper "linking" the circuit. According to Lenz's law the induced cirrent opposes the change. Thus the induced current is counter clock wise. Positive charge builds up on
the bottom plate.
Motional EMF
Part 3
If the moving moving conductor is part of a circuit, then the magnetic flux through the circuit will change with time and a current will be induced (Area of loop = ls):
s
x
x
x
x
x
xx
x
x x
x
x
xx
x
x
x
x
v l
R
(1) l s
N B Blv t t
ε ΔΦ Δ
= = = Δ Δ
Problem The figure shows a zero-resistance rod sliding to the right on two zero-resistance rails separated by the distance L = 0.45 m. The rails are connected by a 12.5 Ω resistor, and the entire system is in a uniform magnetic field with a magnitude of 0.75 T.
(a) If the velocity of the bar is 5.0 m/s to the right, what is the current in the circuit? (b) What is the direction of the current in the circuit? (c) What is the magnetic force on the bar? (d) What force must be applied to keep the bar moving at constant velocity?
0.75 0.45 5 a) or 0.14 A
12.5 Blv
emf Blv IR I R
× × = = = = =
b) clockwise
c) to the left .
0.75 0.45 0.14 0.046 N
F BlI
F
=
= × × =
r
r
d) A force must be applied to the right equal to 0.046 N.
5
Motional emf An emf will also be produced if a conductor moves through a magnetic field. The emf comes from the motion of charges, which are free to move in the conductor. In this example, why do positive charges collect at the top of the rod?
x
x
x
x
x
xx
x
x x
x
x
xx
x
x
x
x
v L
+ +
--
Computer Simulation
Generators A generator is a device that converts mechanical energy to electrical energy. Consider a current loop which rotates in a constant magnetic field: The magnetic flux through the loop changes, so an emf is induced. If a loop of area A with N turns rotates with angular speed ω (period of rotation T = 2π/ω) in a constant B field, then the instantaneous induced emf is:
If this loop is part of a circuit, this emf will induce an Alternating Current (AC) in the circuit.
tNBA ωωε sin=
Generator
A coil of wire turns in a magnetic field. The flux in the coil is constantly changing, generating an emf in the coil.
Transformers A transformer is a device used to change the voltage in a circuit. AC currents must be used.
s
p
s
p
p
s N N
V V
I I
== p = primary
s = secondary
Step-down transformers 240,000 V in the power lines
120 V in houses
2,400 V local substations
6
Eddy Currents When a conductor is moved in a magnetic field, there is a force on the electrons (remember they are free to move in a conductor), which then move in the metal. This movement is called an eddy current.
The induced currents produce magnetic fields which tend to oppose the motion of the metal.
××
×
×
×
× ×
×
×
×
×
× ×
× ×
× × ×
× ×
Self-inductance
Part 4
Solenoids
If we stack several current loops together we end up with a solenoid. In the limit of a very long solenoid, the magnetic field inside is very uniform:
B=μ0nI n = number of windings per unit length, I = current in windings
Self-Inductance
If you try to change the current instantaneously in a circuit containing a solenoid, the response will instead be gradual. This is because the changing magnetic flux through the solenoid produces a self-induced emf to initially oppose any changes as prescribed by Lenz’s Law. This effect is known as self-induction. Actually all circuit elements have some self-inductance. However, only coils with many turns of wire, such as in a solenoid, have substantial self-inductance.
Inductance The self-induced emf is given by:
where L is defined as the inductance of the circuit, usually residing in the “inductor. For any geometry of loop, the magnetic flux through the loop, produced by current in the loop is proportional to the current. The inductance L is the constant of proportionality.
The unit of inductance is the Henry 1 H = 1 T·m2/A = 1 (T·m2/s) (s/A) = 1 V·s/A Note that inductance, like capacitance, is purely geometrical.
t I
L t
N Δ Δ
= Δ ΔΦ
=ε
I NLLIN Δ ΔΦ
=⇒=Φ
Inductance of a Solenoid
A solenoid has inductance given by
l l
AnA N
L 20 2
0 μμ =⎟ ⎟ ⎠
⎞ ⎜ ⎜ ⎝
⎛ =
L = inductance of the solenoid N = number of turns in solenoid l = length of solenoid A = cross sectional area of solenoid n = number turns per length
7
Problem The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross- sectional area of the solenoid? (b) What is the induced emf in the solenoid if its current drops from 3.2 A to 0 in 55 ms?
( )
( )
272 30
3 3 2 27
4 10 450 a) 7.3 10
0.24 0.24
7.3 10 6.88 10 m 4 10 450
AN A L
l
A
πμ
π
− −
− −
−
× × × = = = ×
= × × = × × ×
( )3 3
b) but or
0 3.2 so 7.3 10 0.425 volts
55 10
emf N L N N L I t I
I emf L
t −
−
ΔΦ ΔΦ = − = ΔΦ = Δ
Δ Δ −Δ
= − = − × × = Δ ×
RL Circuits We can construct a circuit out of inductors and resistors. The circuit will behave similar to an RC circuit, with a time constant given by: τ = L/R
)1()1( // LtRt e R
e R
I −− −=−= εε τ
Week 3/Week 3/5.pdf
1
Magnetism
Chapter 20 Part 1
The Magnetic Field
What creates magnetic fields?
There is no magnetic charge!
Any permanent magnet has two poles.
What if to cut a permanent magnet in half?
Unlike electrostatics: Magnetic monopoles have never been detected.
Interaction between magnetic poles
Opposite magnetic poles attract each other, and like poles repel each other
Magnetic Field Lines
Assume that there is a magnetic field in some area of space
We can represent magnetic fields with field lines, as we did for electric fields
(1) the direction of the tangent to a magnetic field line at any point gives the direction of at that point
(2) the spacing of the lines represents the magnitude of
B r
B r
B r
2
Magnetic Field Lines (cont.)
The field lines enter one end of a magnet and exit the other end.
The end of a magnet from which the field lines emerge is called the north pole of the magnet
the other end, where field lines enter the magnet, is called the south pole
The Earth’s Magnetic Field
The spinning iron core of the earth produces a magnetic field.
The magnetic north pole corresponds to the geographic south pole.
South _____
Magnetic Force on a Particle
Part 2
Magnetic force on a charged particle
q – electric charge v – particle velocity B – magnetic field the magnitude
BvqFB rrr
×=
φsinvBqFB =
Right Hand Rule / Teddy Bear Rule / Alligator Rule
Right Hand Rule: Positive and negative particle
BvqFB rrr
×=
The force acting on a charged particle moving through a magnetic field is always perpendicular to the velocity and the field
Magnetic field (definition)
q – electric charge v – particle velocity B – magnetic field
SI unit: Tesla 1 T = newton/(C*m/s) = 1 N/(A*m) 1 tesla = 104 gauss (G)
vq F
B B=
φsinvBqFB =
3
Notation
To depict a vector oriented perpendicular to the page we use crosses and dots.
A cross indicates a vector going into the page (think of the tail feathers of an arrow disappearing into the page).
A dot indicates a vector coming out of the page (think of the tip of an arrow coming at you out of the page).
B out of the page B into the page
×·
Direction of Magnetic Forces (cont.) The direction of the magnetic force on a moving charge is perpendicular to the plane formed by B and v.
F
Β
v
To determine the direction, you must apply the Right Hand Rule (RHR).
Good to discuss
A magnetic field exerts a force on a charged particle: A) always B) never C) if the particle is moving across the field lines D) if the particle is moving along the field lines E) if the particle is at rest
φsinvBqFB =
? Example A charge of 23 mC is moving in the negative x direction at 4 m/s. A magnetic field of 30 T is pointing in the positive y direction. What is the magnitude and direction of the force on the charge? How does your answer change if the charge is –23 mC?
6
3
sin 23 10 4 30 sin 90
2.76 10 N
F qvB
F
θ −
−
= = × × × ×
= ×
o r
r
Use the right-hand rule!
ˆDirection is , i.e. into the paper
If the charge is negative,
ˆ Direction is .
z
z
−
problem
Part 3
The Motion in a Magnetic Field
Motion of Charges in B Fields
If a charged particle is moving in a direction perpendicular to a uniform magnetic field, then its trajectory will be a circle because the force F=qvB is always perpendicular to the velocity, and therefore centripetal.
r mv
maFc 2
== r vm
qvBF 2
==
qB vm
r =
Recall that so
The radius of the circular trajectory
The period (the time for one full revolution) qB m
v r
T ππ 22
==
4
Good to discuss
An electron and a proton are both initially moving with the same speed and in the same direction at 900 to the same uniform magnetic field. They experience magnetic forces, which are initially: A) identical B) equal in magnitude but opposite in direction C) in the same direction and differing in magnitude by a factor of 1840 D) in opposite directions and differing in magnitude by a factor of 1840 E) equal in magnitude but perpendicular to each other
F qvB mv r
= = 2 ?
Good to discuss
An electron and a proton each travel with equal speeds around circular orbits in the same uniform magnetic field, as shown in the diagram (not to scale). The field is into the page on the diagram. (a) Where is the electron (b) What is the direction
qB mv
r =
?
Good to discuss A uniform magnetic field is directed into the page. A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown. A reasonable explanation is: A) the charge is positive and slowing down B) the charge is negative and slowing down C) the charge is positive and speeding up D) the charge is negative and speeding up E) none of the above
qB mv
r =
? Good to discuss
Cosmic rays (atomic nuclei stripped bare of their electrons) would continuously bombard Earth’s surface if most of them were not deflected by Earth’s magnetic field. Given that Earth is, to an excellent approximation, a magnetic dipole (a bar magnet), the intensity of cosmic rays bombarding its surface is greatest at the 1. poles. 2. mid-latitudes. 3. equator.
?
Isotope Separation qB mv
r = Crossed E and B fields
5
Magnetic Force on a Current
Part 4
Force on a Current Carrying Wire Current in a wire is a collection of moving charges; therefore, a current carrying wire in a magnetic field also experiences a force. If a wire of length L, carrying a current I, makes an angle θ with a magnetic field B, then the magnitude of the force on the wire is:
θ Ι
Β
• F
θϑ
ϑϑ
sinsin)(
sin)(sin
ILBvB v L
I
vBItqvBF
==
=== θsinILBF =
Force on a Current Carrying Wire θsinILBF =
Force on a Current Carrying Wire θsinILBF = The diagram shows a straight wire carrying a flow of electrons into the page. The wire is between the poles of a permanent magnet. The direction of the magnetic force exerted on the wire is: A) → B) ← C) ↓ D) ↑ E) into the page
?
Magnetic Levitation (Maglev, etc.)
mgILB=
Magnetic Torque on current loop
In a uniform magnetic field, the net force on a current loop (independent of geometry) is 0. However, there can be a torque
r = distance from axis of rotation to loop segment F = magnetic force on segment θ = angle between vector r and vector F.
Only the vertical segments of the loop experience a force. The torque will rotate the loop so that the plane of the loop is perpendicular to the magnetic field.
θτ sinrF=
)( 22
hwIB w
IhB w
IhB =+=τ
IBA=τ
6
Magnetic Torque on current loop (cont.)
Torque exerted on a rectangular loop of area A
If the loop has N turns, then
θτ sinIBA=
θτ sinNIBA=
A square loop of wire lies in the plane of the page and carries a current I as shown. There is a uniform magnetic field parallel to the side MK as indicated. The loop will tend to rotate: A) about PQ with KL coming out of the page B) about PQ with KL going into the page C) about RS with MK coming out of the page D) about RS with MK going into the page E) about an axis perpendicular to the page
?
Magnetic Fields Due to Currents
Part 5
Experimental observation in 1820
Hans Oersted:
Electric currents can create magnetic fields
The magnitude of the field produced at point P
A general equation
for Physics 232
AmT r
ids dB
/1026.1
sin 4
6 0
2 0
⋅×=
=
−μ
ϑ π μ
A Long Straight Wire
Proof: integration or Ampere’s law
AmT R i
B
/1026.1 2
6 0
0
⋅×=
=
−μ π μ
7
Conceptual question The equation above is true for an infinitely long, straight conductor carrying a current.
Of course, there is no such thing as an infinitely long anything. How would you decide whether a particular wire is long enough to be considered infinite?
? R i
B π μ 2
0= Problem
Consider the long, straight, current-carrying wires shown in the figure. One wire carries a current of 6.2 A in the negative y direction; the other carries a current of 4.5 A in the positive x direction. Calculate the magnitude and direction of the net magnetic field at points A and B.
problem
Force between two parallel currents
Magnitude of Ba
d i
B aa π μ 2
0=
Force on a length L of wire b
abba LBiF =
d iLi
F baba π μ
2 0=
Each of two parallel wires with currents I1 and I2, experiences a magnetic force given by
L = length of wire d = distance between the two wires
If the currents are parallel, the force is attractive. If the currents are anti-parallel the force is repulsive.
L d II
F π
μ 2
210=
Good to discuss Two long parallel straight wires carry equal currents in opposite directions. At a point midway between the wires, the magnetic field they produce is: A) zero B) non-zero and along a line connecting the wires C) non-zero and parallel to the wires D) non-zero and perpendicular to the plane of the two wires E) none of the above
? Good to discuss
On a computer chip, two conducting strips carry charge from P to Q and from R to S. If the current direction is reversed in both wires, the net magnetic force of strip 1 on strip 2 1. remains the same. 2. reverses. 3. changes in magnitude, but not in direction. 4. changes to some other direction. 5. other
?
8
Conceptual question Streams of charged particles emitted from the sun during unusual sunspot activity create a disturbance in the earth’s magnetic field (called a magnetic storm). How can they cause such a disturbance?
? B Fields of Current Distributions By winding wires in various geometries, we can produce different magnetic fields. For example, a current loop (perpendicular to plane, radius R, current emerging from plane at top of loop):
Ι
Magnitude of magnetic field at the center of loop:
R IN
B 2
0μ=
N = # of loops of wire (i.e. # turns)
Direction of magnetic field from the RHR.
Solenoids
If we stack several current loops together we end up with a solenoid. In the limit of a very long solenoid, the magnetic field inside is very uniform:
n = number of windings per unit length, I = current in windings
nIB 0μ=
Magnetic Materials
On atomic level – moving electrons (microscopic current loops) create magnetic fields. In many materials, these currents are randomly oriented (net magnetic field is zero). In some materials, the presence of an external magnetic field can cause the loops to become oriented
Paramagnetism – orientation with an external field Diamagnetism – orientation against an external field Ferromagnetism – line up loops (magnetic domains) without an external field
