Concept of magnetic field

Motivating the Magnetic Field Concept and magnetic field conceptual questions
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Dr.TomHunt,United States,Teacher
Published Date:22-07-2017
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UNIT - III UNIT - III Static Magnetic Fields Static Magnetic Fields B.Hemalatha AP-ECE 1Motivating the Magnetic Field Concept: Motivating the Magnetic Field Concept: Forces Between Currents Forces Between Currents Magnetic forces arise whenever we have charges in motion. Forces between current- carrying wires present familiar examples that we can use to determine what a magnetic force field should look like: Here are the easily-observed facts: B.Hemalatha AP-ECE 2Biot-Savart Law Biot-Savart Law • The current flowing in any path, can be considered as many infinitesimal current elements, each of length dl, flowing in a magnetic field dB, at any point P ˆ Idlr 0 dB 2 4r B.Hemalatha AP-ECE 3Idl sin 0 • The magnitude of dB is: dB 2 4r where θ is the angle between dl and r ˆ Idlr 0 BdB • The total magnetic field at point P is: 2  2r • This is equivalent to Coulomb’s Law written in differential form: 1dq dE 2 4r 0 B.Hemalatha AP-ECE 4Magnetic Field due to a Current- Carrying Wire Biot-Savart Law Hans Christian Oersted, 1820 B.Hemalatha AP-ECE 5• Magnetic fields are caused by currents. • Hans Christian Oersted in 1820’s showed that a current carrying wire deflects a compass. No Current in the Wire Current in the Wire B.Hemalatha AP-ECE 6Right Hand Curl Rule Right Hand Curl Rule B.Hemalatha AP-ECE 7Magnetic Fields of Long Current-Carrying Wires B =  I o 2r I I = current through the wire (Amps) r = distance from the wire (m)  = permeability of free space o -7 = 4 x 10 T m / A B = magnetic field strength (Tesla) B.Hemalatha AP-ECE 8Magnetic Field of a Current Carrying Wire B.Hemalatha AP-ECE 9What if the current-carrying wire is not straight? Use What if the current-carrying wire is not straight? Use the Biot-Savart Law: the Biot-Savart Law: Assume a small segment of wire ds causing a field dB: ˆ  dsr  0 dBI 2  4  r Note: dB is perpendicular to ds and r B.Hemalatha AP-ECE 10Biot-Savart Law allows us to calculate the Biot-Savart Law allows us to calculate the Magnetic Field Vector Magnetic Field Vector • To find the total field, sum up the contributions from all the current elements I ds ˆ  dsr  0ii BI  2  4  r i • The integral is over the entire current distribution ˆ μIdsr o B 2  4πr B.Hemalatha AP-ECE 11Note on Biot-Savart Law Note on Biot-Savart Law • The law is also valid for a current consisting of charges flowing through space • ds represents the length of a small segment of space in which the charges flow. • Example: electron beam in a TV set B.Hemalatha AP-ECE 12Comparison of Magnetic to Electric Field Magnetic Field Electric Field 2 2 • B proportional to r • E proportional to r • Vector • Vector • Perpendicular to F , ds, r • Same direction as F B E • Magnetic field lines have no • Electric field lines begin on beginning and no end; they positive charges and end on form continuous circles negative charges • Biot-Savart Law • Coulomb’s Law • Ampere’s Law (where there • Gauss’s Law (where there is is symmetry symmetry) B.Hemalatha AP-ECE 13Derivation of B for a Long, Straight Current-Carrying Derivation of B for a Long, Straight Current-Carrying Wire Wire Integrating over all the current elements gives ˆ ˆ dsrdx sin θ k  θ μI 2 o Bsinθ dθ  θ 1 4πa μI o cosθcosθ  1 2 4πa B.Hemalatha AP-ECE 14If the conductor is an infinitely long, straight If the conductor is an infinitely long, straight wire, = 0 and =  wire, = 0 and =    • The field becomes: a μI o B 2πa B.Hemalatha AP-ECE 15B.Hemalatha AP-ECE 16B for a Curved Wire Segment B for a Curved Wire Segment • Find the field at point O due to the wire segment A’ACC’: B=0 due to AA’ and CC’ Due to the circular arc: ˆ μIdsr o B 2  4πr μI o Bθ 4πR q=s/R, will be in radians B.Hemalatha AP-ECE 17B at the Center of a Circular Loop of Wire B at the Center of a Circular Loop of Wire μIμI oo Bθ 2π 4πR 4πR Consider the previous result, with  = 2 μI o B 2R B.Hemalatha AP-ECE 18Note The overall shape of the magnetic field of the circular loop Is similar to the magnetic field of a bar magnet. B.Hemalatha AP-ECE 19B along the axis of a Circular Current Loop B along the axis of a Circular Current Loop • Find B at point P ˆ μIdsr o B 2  4πr 2 μIR o If x=0, B same as at center of a loop B x 3 2 2 2 2xR  B.Hemalatha AP-ECE 20

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