Lecture 14: TUE 09 MARLecture 14: TUE 09 MARMagnetic Fields IIMagnetic Fields II

Ch.28.6-10

Physics 2102

Jonathan Dowling

“They are not supposedto exist….”

Aurora Borealis

Second Exam Review:10:40AM–12:00N THU 11 MARNicholson 109 in Class

Second Exam (Chapters 24–28):6–7PM THU 11 MARLockett 6

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

B into blackboard.

vF

FB = FC

! qv B = mv 2

r

In general, path isa helix (component of v parallelto field is unchanged).

r

Fcentrifugalout = ma = mr! 2 = m v 2

r

Fmagneticin = qvB

Solve : r = mvqB

.

.electron

!B

v!

F!

Cr

r = mvqB

! = vr

= qBm

T ! 2"rv

= 2"mvqBv

= 2"mqB

f ! 1T

= qB2"m

Radius of Circlcular Orbit

Angular Frequency:Independent of v

Period of Orbit:Independent of v

Orbital Frequency:Independent of v

ExampleExampleTwo charged ions A and B traveling witha constant velocity v enter a box inwhich there is a uniform magnetic fielddirected out of the page. Thesubsequent paths are as shown. Whatcan you conclude?

qBmvr =

Same charge q, speed v, and same B for both masses. So: ion with larger mass/charge ratio (m/q) moves in circle of largerradius. But that’s all we know! Don’t know m or q separately.

(a) Both ions are negatively charged.

(b) Ion A has a larger mass than B.

(c) Ion A has a larger charge than B.

(d) None of the above.

v

v

A

B

Aurora borealis(northern lights)

Synchrotron

Linear accelerator (long).

Fermilab,Batavia, IL (1km)

Suppose you wish to accelerate chargedparticles as fast as you can.

Examples of Circular Motion inExamples of Circular Motion inMagnetic FieldsMagnetic Fields

Magnetic Force on a WireMagnetic Force on a Wire

dvLitiq ==

LBvqF d

!!!!=

BLiBqLiqF

!!!!

!!=!=

BLiF!!!

!=

BLdiFd!!!

!=

Note: If wire is not straight,compute force on differentialelements and integrate:

ExampleExample

iLBFF == 31

!iBRdiBdLdF ==

By symmetry, F2 will only have a vertical component,

iBRdiBRdFF 2)sin()sin(00

2 !! ===""

###

)(22321total RLiBiLBiRBiLBFFFF +=++=++=Notice that the force is the same as that for a straight wire,

L LR Rand this would be true nomatter what the shape of the central segment!.

Wire with current i.Magnetic field out of page.What is net force on wire?

Example 4: The Rail GunExample 4: The Rail Gun• Conducting projectile of length

2cm, mass 10g carries constantcurrent 100A between two rails.

• Magnetic field B = 100T pointsoutward.

• Assuming the projectile startsfrom rest at t = 0, what is itsspeed after a time t = 1s?

B I L

• Force on projectile: F= iLB (from F = iL x B)• Acceleration: a = F/m = iLB/m (from F = ma)• v = at = iLBt/m (from v = v0 + at)

= (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s= 4,473mph = MACH 8!

projectile

rails

Rail guns in the “Eraser” movie"Rail guns are hyper-velocity weapons that shoot aluminum or clay rounds atjust below the speed of light. In our film, we've taken existing stealthtechnology one step further and given them an X-ray scope sighting system,"notes director Russell. "These guns represent a whole new technology inweaponry that is still in its infancy, though a large-scale version exists inlimited numbers on

battleships and tanks. They haveincredible range. They can piercethree-foot thick cement walls andthen knock a canary off a tin canwith absolute accuracy. In our film,one contractor has finally developedan assault-sized rail gun. Weresearched this quite a bit, and thetechnology is really just around thecorner, which is one of the excitingparts of the story."

Warner Bros., production notes, 1996.http://movies.warnerbros.com/eraser/cmp/prodnotes.html#tech

Also: INSULTINGLY STUPID MOVIE PHYSICS: http://www.intuitor.com/moviephysics/

Rail guns in Transformers II

Electromagnetic Slingshot

These Devices CanLaunch 1000kg ProjectilesAt Mach 100 at a Rate of1000 Projectiles Per Second.

Using KE = 1/2mv2

This corresponds to anoutput about 1012 Watts =TeraWatt.

Uses: Put Supplies on Mars.Destroy NYC in about 10minutes.

Torque on a Current Loop: Principle behind electric motors.

Net force on current loop = 0

iaBFF == 31

)sin(1 !FF ="

)sin(!" iabBbFTorque === #

For a coil with N turns,τ = N I A B sinθ, where A is the area of coil

Rectangular coil: A=ab, current = i

But: Net torque is NOT zero!

nNiA ˆ)(=µ! n̂,µ!

Magnetic Dipole MomentMagnetic Dipole Moment

N = number of turns in coilA=area of coil.

We just showed: τ = NiABsinθRight hand rule:curl fingers in

direction of current;thumb points along µDefine: magnetic dipole moment µ

B!!! != µ"

As in the case of electric dipoles, magnetic dipoles tend to alignwith the magnetic field.

Electric vs. Magnetic DipolesElectric vs. Magnetic Dipoles

UE = ! ! p "! E

-Q

θQE

QE

+Q

p=Qa

! ! B = ! µ "! B

nNiA ˆ)(=µ!

! ! E = ! p "! E

UB = !! µ "! B

Magnetic Force on a Wire.Magnetic Force on a Wire.

L

BLiF!!!

!= BLdiFd!!!

!=

. dL!

B!

dF!

i!

If we assume the more general case for which the

magnetic field forms an angle with the wirethe magnetic force equation can be written

B !

Magnetic Force on a Straight Wire in a Uniform Magnetic Field

!

in vector

form as . Here is a vector whose magnitude is equal to the wire length and has a direction that coincides with that of the current.The magnetic force magnitude is s

B

B

F iL B LL

F iLB

= "

=

! ! ! !

in .

In this case we divide the wire into elements oflength , which can be considered as straight.The magnetic force

dL

!Magnetic Force on a Wire of Arbitrary ShapePlaced in a Nonuniform Magnetic Field

on each element is

. The net magnetic force on the

wire is given by the integral .B

B

dF idL B

F i dL B

"

= "#=

! ! !

! ! !

BF iL B= !! ! !

BdF idL B!=! ! !

BF i dL B= !"! ! !

(28-12)

Torque on a Current Loop: Principle behind electric motors.

Net force on current loop = 0

iaBFF == 31

)sin(1 !FF ="

)sin(!" iabBbFTorque === #

For a coil with N turns,τ = N I A B sinθ, where A is the area of coil

Rectangular coil: A=ab, current = i

But: Net torque is NOT zero!

nNiA ˆ)(=µ! n̂,µ!

Magnetic Dipole MomentMagnetic Dipole Moment

N = number of turns in coilA=area of coil.

We just showed: τ = NiABsinθRight hand rule:curl fingers in

direction of current;thumb points along µDefine: magnetic dipole moment µ

B!!! != µ"

As in the case of electric dipoles, magnetic dipoles tend to alignwith the magnetic field.

Electric vs. Magnetic DipolesElectric vs. Magnetic Dipoles

UE = ! ! p "! E

-Q

θQE

QE

+Q

p=Qa

! ! B = ! µ "! B

nNiA ˆ)(=µ!

! ! E = ! p "! E

UB = !! µ "! B