lec23_030207_-_Magnetics_IV

# lec23_030207_-_Magnetics_IV - Magnetics IV Magnetic Force...

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Unformatted text preview: Magnetics IV Magnetic Force: Big Picture dFmag = Id B Fmag = qv B 4 ^ o Id r r2 = 4 Biot-Savart Law: dB = Biot- o Id r r3 B-field of a finite wire: B = - For a = b: B = - o I 2x o I 4x ^ k Electrostatics Circuits Magnetics Chapter 27: Magnetic Force on Charges and currents Chapter 28: Sources of Magnetic Field Chapter 29: Electromagnetic Induction Chapter 30: Self Inductance and Inductors a a +x 2 2 + y d b b + x2 2 ^ k a a2 + x2 For infinite wires: B = - o I ^ k 2x a b r P B Hall Effect: nq = - J x By Ez =- J x By d Vz I x Optics 1 Friday, 2 March 2007 2 Friday, 2 March 2007 B-field on the Axis of a Current Loop I B-field on the Axis of a Current Loop I d R y ^ d = - Rd sin ^ + Rd cos k j r P d y B x ^ ^ r = x i - R cos ^ - R sin k j R r P B x B = dB = o I d r 4 r3 Id r dB = o 4 r 3 dB = ^ ^ o I - Rd sin ^ + Rd cos k x i^ - R cos ^ - R sin k j j 4 2 2 3/ 2 ( 2 2 2 2 ^ j I (xRd sin ) k + (R d sin ) i^ + (xRd cos ) ^+ R d cos i^ = o 3/ 2 4 x2 + R2 (x ) ( ) ) ) dB = +R ^ ^ j o I (R 2 d sin 2 + R 2 d cos 2 ) i + (xRd cos ) ^ + (xRd sin ) k 3/ 2 4 (x 2 + R 2 ) ( ) ( B= o I 4 2 0 (x ^ R 2 d i 2 +R 2 3/ 2 ) = 2R 2 o I i^ 4 (x 2 + R 2 )3 / 2 = IA = IR 2 o 2 B= Friday, 2 March 2007 3 Friday, 2 March 2007 4 x 2 + R 2 ( ) 3/ 2 4 1 Question In the limit that x >> R, the magnetic field due to a R, o 2 current loop B = becomes: 2 2 3/ 2 4 x + R Question At x = 0, the magnetic field due to a current loop 0, o 2 B= 2 2 3 / 2 becomes: 4 x + R A. o 23 4R x ( ) ( ) A. Zero B. C. D. 5 B. 4x 3 o 2 o I 2R C. 4 x 3 D. o 2 o IR 2 2 o 2 4Rx 3 o I 2R 3 6 Friday, 2 March 2007 Friday, 2 March 2007 Outside a magnet, the magnetic field point From the North Pole to the South Pole! y I Magnetic Field Outside a magnet, the magnetic field point From the North Pole to the South Pole! Magnetic Field I-Carrying Loop: R B N x S Solenoid: S N Permanent Magnet: B Friday, 2 March 2007 7 Friday, 2 March 2007 8 2 How do we calculate the magnetic field of a solenoid? By adding up the field due to a large number of current loops! N total turns, Current I n = N/L = loop number density For one loop, we have: B1 loop = dBx = Solenoids Solenoids For N loops, we multiply by N. But these loops are not all at the same distance x! INTEGRATE! to find Bx at x = 0! Friday, 2 March 2007 o I 2R 2 4 (x 2 + R 2 )3 / 2 o 2R 2 Indx 4 (x 2 + R 2 )3 / 2 Bx = dBx = o 2 b nIR 2 -a (x dx 2 + R2 ) 3/ 2 Bx = 9 Friday, 2 March 2007 N = nL dN = ndL = ndx 1 o nI 2 b b +R 2 2 + a a + R2 2 10 In the limit that a >> R and b >> R, the magnetic R, 1 b a field for a solenoid Bx = o nI + 2 b2 + R2 a 2 + R2 becomes: A. Zero B. C. D. E. Friday, 2 March 2007 Question Quiz At one end of the solenoid, a = 0 and b b a magnetic field B = 1 nI + x 2 o , the becomes: b2 + R2 a 2 + R2 A. Zero B. C. D. E. 11 Friday, 2 March 2007 1 o nI 2 1 o nI 2 o nI 2 o nI o nI 2 o nI None of the above None of the above 12 3 "Gauss's Law" for Magnetism Gauss' Law" For electric field ... d I R Ampere's Law Ampere' B d = o I thru E = E dA +q -q E dA = Qencl B o What about for magnetics? magnetics? By symmetry, the magnetic field lines will be circular around the wire, as is the d : B d = B d = o I thru By symmetry, the magnetic field will have the same strength all along the path: B d = B d = B d = o I thru B = B dA B dA = Friday, 2 March 2007 QB , encl o ? B dA = 0 13 B d = o I thru Friday, 2 March 2007 B(2R ) = o I thru B= o I thru 2R 14 Example: Long Solenoid 3 Example: Long Solenoid Bd = Bd + Bd + Bd + Bd 1 3 B d = o I encl 2 2 3 4 Bd = 0 1 2 4 1 d Bin 4 1. Choose a path. 1 d Bout Bfar out = 0 2 Bd = out Bd + Bd Bout d in 2. Left hand side of Ampere's law. Ampere' 1. Choose a path. Bd = Bd + Bd + Bd + Bd 1 2 3 4 2. Left hand side of Ampere's law. Ampere' =0 Bin d Bd = Bd + Bd 4 in out 16 15 Friday, 2 March 2007 Friday, 2 March 2007 =0 4 Example: Long Solenoid Bd = Bd + Bd + Bd + Bd 1 3 Example: Long Solenoid o I encl = o (nd )I B d = B d = Bin d 3 3 2 3 4 Bd = Bd 3 2 B d = o I encl 2 Bin 4 Bin 1 d = Bin d 3 4 Bin d = o (nd )I Bin = o nI 1 d 1. Choose a path. 2. Left hand side of Ampere's law. Ampere' 3. Right hand side of Ampere's law. Ampere' = Bin d 3 1. Choose a path. 2. Left hand side of Ampere's law. Ampere' 3. Right hand side of Ampere's law. Ampere' 4. Ampere's law! Ampere' 17 Friday, 2 March 2007 = Bin d number of turns/length o I encl = o I total = o NI = o (nd )I Friday, 2 March 2007 number of turns 18 5 ...
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## This note was uploaded on 04/08/2008 for the course PHGN 200 taught by Professor Japguy during the Fall '07 term at Mines.

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