Lecture 25: WED 21 OCT - LSUjdowling/PHYS21133-FA15/lectures/25WED21OCT.pdfLecture 25: WED 21 OCT...

Lecture 25: WED 21 OCT Magnetic Fields III

Physics 2113 Jonathan Dowling

“They are not supposed to exist….”

Aurora Borealis

Magnetic Force on a Wire.

L

BLiF!!!

×= BLdiFd!!!

×=

Magnetic Force on a Wire

dvLitiq ==

L BvqF d

!!!×=

BLiBqLiqF

!!!!

!×=×=

BLiF!!!

×=

BLdiFd!!!

×=

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

i

. dL!

B!

dF!

i φ

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 φ

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

φMagnetic 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)

i

28.8.2. A portion of a loop of wire passes between the poles of a magnet as shown. We are viewing the circuit from above. When the switch is closed and a current passes through the circuit, what is the movement, if any, of the wire between the poles of the magnet?

a) The wire moves toward the north pole of the magnet. b) The wire moves toward the south pole of the magnet. c) The wire moves upward (toward us). d) The wire moves downward (away from us into board). e) The wire doesn’t move.

Example

iLBFF == 31

θiBRdiBdLdF ==

By symmetry, F2 will only have a vertical component,

iBRdiBRdFF 2)sin()sin(00

2 ∫∫ ===ππ

θθθ

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

L L R R and this would be true no matter what the shape of the central segment!.

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

i−y direction ↓

i

Fgrav = mg Fmag = iLB

To balance:Fgrav = Fmag

⇒ mg = iLB

⇒ B = mgiL

= mL

⎛⎝⎜

⎞⎠⎟gi

m / L = 46.6 ×10−3kg/m

SP28-06

ICPP: Find Direction of B

B = 46.6 × 10–3 kg m | 9.8 m

s2 | s28 C

= 1.6 × 10–2 kg s⋅C

= 16 × 10–3 T = 16 mT

Example 4: The Rail Gun •  Conducting projectile of length

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

•  Magnetic field B = 100T points outward.

•  Assuming the projectile starts from rest at t = 0, what is its speed 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 at just below the speed of light. In our film, we've taken existing stealth technology one step further and given them an X-ray scope sighting system," notes director Russell. "These guns represent a whole new technology in weaponry that is still in its infancy, though a large-scale version exists in limited numbers on

battleships and tanks. They have incredible range. They can pierce three-foot thick cement walls and then knock a canary off a tin can with absolute accuracy. In our film, one contractor has finally developed an assault-sized rail gun. We researched this quite a bit, and the technology is really just around the corner, which is one of the exciting parts 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

Rail guns in Reality!

Electromagnetic Slingshot

These Devices Can Launch 1000kg Projectiles At Mach 100 at a Rate of 1000 Projectiles Per Second. Using KE = 1/2mv2 This corresponds to an output about 1012 Watts = TeraWatt. Uses: Put Supplies on Mars 80 days after launch. Current time to Mars on Ordinary Rocket? 1.5-3 years.

Torque on a Current Loop: Principle behind electric motors.

Net force on current loop = 0

iaBFF == 31

)sin(1 θFF =⊥

)sin(θτ 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 Moment

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

We just showed: τ = NiABsinθ Right hand rule:

curl fingers in direction of current;

thumb points along µ Define: magnetic dipole moment m

B!!! ×= µτ

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

Electric vs. Magnetic Dipoles

�

UE = − ! p ⋅! E

-Q

qQE

QE

+Q

p=Qa

�

! τ B = ! µ ×! B

nNiA ˆ)(=µ!

�

! τ E = ! p ×! E

�

UB = −! µ ⋅! B

nNiA ˆ)(=µ! n̂,µ!

Magnetic Dipole Moment

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

We just showed: t = NiABsinq Right hand rule:

curl fingers in direction of current;

thumb points along µ Define: magnetic dipole moment m

B!!! ×= µτ

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

�

UE = − ! p ⋅! E

!τ =!µ ×!B

τ = µBsinθτ1 = τ 2 = τ 3 = τ 4τ is biggest when B is at right angles to μ

1 and 3 are “downhill”. 2 and 4 are “uphill”. U1 = U4 > U2 = U3

ICPP