lecture14

# lecture14 - Physics 2102 Aurora Borealis Jonathan Dowling...

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Physics 2102 Physics 2102 Lecture 14 Lecture 14 Ch28: Magnetic Forces on Current Wires Ch28: Magnetic Forces on Current Wires Physics 2102 Jonathan Dowling Star Quake on a Magnetar! “I’ll be back…. Aurora Borealis

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Crossed Fields E vs. B Crossed Fields E vs. B F E =qE B E y q F B =vqBF E v L F E =ma => y=qEL 2 /(2mv 2 ) F B =F E => y=0 => v=E/B
Magnetic force on a wire. Magnetic force on a wire. d v L i t i q = = L B v q F d r r r × = B L i B q L i q F r r r r r × = × = B L i F r r r × = B L d i F d r r r × = Note: If wire is not straight, compute force on differential elements and integrate:

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Example iLB F F = = 3 1 θ iBRd iBdL dF = = By symmetry, F 2 will only have a vertical component, iBR d iBR dF F 2 ) sin( ) sin( 0 0 2 = = = π ) ( 2 2 3 2 1 total R L iB iLB iRB iLB F F F F + = + + = + + = Notice that the force is the same as that for a straight wire, L L R R Wire with current i. Magnetic field out of page. What is net force on wire?
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## This note was uploaded on 11/18/2011 for the course PHYSICS 2102 taught by Professor Dowling during the Fall '10 term at LSU.

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lecture14 - Physics 2102 Aurora Borealis Jonathan Dowling...

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