Chapter 29b - Chapter 29 Magnetic Fields Due to Currents...

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Chapter 29: Magnetic Fields Due to Currents Calculating the magnetic field with Biot-Savart law. HITT question #1. Ampere’s Law. HITT question #2. Magnetic field outside a long straight current carrying wire. Magnetic field inside a long straight current carrying wire . October 14, 2009 Ch. 29: Magnetic Fields Due to Currents - Part B 1
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Magnetic Fields due to Currents Biot-Savart law. ˆ s r 0 3 4 ids r dB r 1 0 m / A B at a perpendicular distance R from a long 7 0 41 Tm pp g (infinite) straight wire carrying current i . 0 2 i B R October 14, 2009 Ch. 29: Magnetic Fields Due to Currents - Part B 2
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Magnetic Fields due to Currents B at a perpendicular distance R from a finite straight wire with length L carrying current i . Proof by Biot-Savart law. All dB vectors are into the page. Integrate dB for s from 0 to L . 0 sin ids B 2 4 dB r 00 sin LL L ii R d B d s d s 22 2 3 2 0 44 / () Bd rs R    0 L L   October 14, 2009 Ch. 29: Magnetic Fields Due to Currents - Part B 3 1 2 1 2 0 24 / / ( ) sL B RR sR L R   
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Magnetic Fields due to Currents The magnitude of the magnetic field B at the center of an arc of radius R and carrying current i . 0 4  i B R B The force between parallel currents – the force on wire b due to the magnetic field produced by the current in wire a . Parallel currents attract each other, anti-parallel currents repel each other. ab ba Li i F 0 October 14, 2009 Ch. 29: Magnetic Fields Due to Currents - Part B 4 d 2
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HITT Question 1 The figure shows three long, straight, parallel, equally spaced wires with identical currents. The agnitude of the force on each wire due to the magnitude of the force on each wire due to the currents in the other two wires is F 1 , F 2 and F 3 , spectively. Which statement is true? respectively. Which statement is true? A) F 3 >F 2 >F 1 ; B) F 2 >F 1 >F 3 ; C) F 2 >F 3 >F 1 ; ) =F F October 14, 2009 Ch. 29: Magnetic Fields Due to Currents - Part B 5 D) F 2 F 3 >F 1 ; E) F 1 =F 3 >F 2 .
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HITT Question 1 Solution A) F 3 >F 2 >F 1 ; B) F 3 >F 1 >F 2 ; C) F 2 >F 3 >F 1 ; D) F 2 =F 3 >F 1 ; ) =F >F E) F 1 F 3 F 2 . 2 0 11 21 3 ˆ 2 Li FF F i d     22 dd  2 0 2 1 2 3 ˆ Li F i  2 2 0 ˆ Li F F i  October 14, 2009 Ch. 29: Magnetic Fields Due to Currents - Part B 6 33 1 3 2
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HITT Question 1 The figure shows three long, straight, parallel, equally spaced wires with identical currents. The agnitude of the force on each wire due to the
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Chapter 29b - Chapter 29 Magnetic Fields Due to Currents...

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