Lecture 25: FRI 24 OCT Magnetic Fields III

Transcription

1 Physics 2113 Aurora orealis Jonathan Dowling Lecture 25: FRI 24 OCT Magnetic Fields III They are not supposed to exist.

2 Magnetic Force on a Wire. L F = i L df = i dl

3 Magnetic Force on a Wire v d L i i t q = = L qv F d = L i q L i q F = = L i F = dl i df = Note: If wire is not straight, compute force on differential elements and integrate: i

4 F = il df φ. i dl df = idl F = i dl Magnetic Force on a Straight Wire in a Uniform Magnetic Field If we assume the more general case for which the magnetic field forms an angle φ with the wire the magnetic force equation can be written in vector form as F = il. Here L is a vector whose magnitude is equal to the wire length L and has a direction that coincides with that of the current. The magnetic force magnitude is F = ilsin φ. Magnetic Force on a Wire of Arbitrary Shape Placed in a Nonuniform Magnetic Field In this case we divide the wire into elements of length dl, which can be considered as straight. The magnetic force on each element is df = idl. The net magnetic force on the wire is given by the integral F = i dl. (28-12)

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6 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. i

7 π π Example F F = 1 = 3 il irdθ df = idl = F = sin( θ ) df = ir sin( θ ) dθ 2iR = y symmetry, F 2 will only have a vertical component, F = F + F + F = il + 2iR + il = 2i( L + total Notice that the force is the same as that for a straight wire, L R R L Wire with current i. Magnetic field out of page. What is net force on wire? and this would be true no matter what the shape of the central segment. R)

8 y direction i

9 F grav = mg F mag = il m / L = kg/m To balance: i F grav = F mag mg = il = mg il = m L g i

10 Example 4: The Rail Gun Conducting projectile of length 2cm, mass 10g carries constant current 100A between two rails. Magnetic field = 100T points outward. Assuming the projectile starts from rest at t = 0, what is its speed after a time t = 1s? rails I L projectile Force on projectile: F= il (from F = il x ) Acceleration: a = F/m = il/m (from F = ma) v = at = ilt/m (from v = v 0 + at) = (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s = 4,473mph = MACH 8

11 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 ros., production notes, Also: INSULTINGLY STUPID MOVIE PHYSICS:

12 Rail guns in Transformers II

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14 Rail guns in Reality

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16 Electromagnetic Slingshot These Devices Can Launch 1000kg Projectiles At Mach 100 at a Rate of 1000 Projectiles Per Second. Using KE = 1/2mv 2 This corresponds to an output about Watts = TeraWatt. Uses: Put Supplies on Mars.

17 Torque on a Current Loop: Rectangular coil: A=ab, current = i Principle behind electric motors. Net force on current loop = 0 ut: Net torque is NOT zero F F = = 3 ia 1 F = F sin( θ ) 1 Torque τ = F b = iabsin(θ ) = For a coil with N turns, τ = N I A sinθ, where A is the area of coil

18 Magnetic Dipole Moment We just showed: τ = NiAsinθ N = number of turns in coil A=area of coil. Define: magnetic dipole moment m µ, µ = ( NiA) nˆ nˆ τ = µ Right hand rule: curl fingers in direction of current; thumb points along m As in the case of electric dipoles, magnetic dipoles tend to align with the magnetic field.

19 Electric vs. Magnetic Dipoles p=qa +Q µ = ( NiA) nˆ -Q q QE QE τ = E p E U E = p E τ = µ U = µ

20 Magnetic Dipole Moment We just showed: t = NiAsinq N = number of turns in coil A=area of coil. Define: magnetic dipole moment m µ, µ = ( NiA) nˆ nˆ τ = µ U = p E E Right hand rule: curl fingers in direction of current; thumb points along µ As in the case of electric dipoles, magnetic dipoles tend to align with the magnetic field.

21 τ = µ τ = µsinθ 1 and 3 are downhill. 2 and 4 are uphill. U1 = U4 > U2 = U3 τ 1 = τ 2 = τ 3 = τ 4 τ is biggest when is at right angles to μ