HR29_post - Chapter 29 Magnetic Fields due to Currents Will...

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Chapter 29 Magnetic Fields due to Currents Will explore the relationship between an electric current and the magnetic field it produces • In order to find magnetic field generated by current in conductors of various geometries, we follow two approaches – For problems with low symmetry we use the Biot-Savart law in combination with the principle of superposition 1 – Introduce Ampere’s law for problems with high symmetry • These two approaches will be used to explore the magnetic field generated by currents in straight wire, arc wire, wire loop, solenoidal coil and toroidal coil • We will also determine the force between two parallel current- carrying conductors. We will then use this force to define the SI unit for electric current (Ampere)
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Oersted observed that a magnetic needle is deflected from its normal north-south orientation in the presence of a wire that carries an electric current. Thus he concluded that an electric current produces in its vicinity a magnetic field 2 Hans Christian Ørsted 1777 – 1851
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• Consider a wire of arbitrary shape carrying current i • Find the magnetic field B created by the wire at a point P . First, we have to consider a small differential element of the wire, ds , and the corresponding current-length element id s . The field d B produced at point P by this current- The law of Biot-Savart 3 length element turns out to be: 2 0 4 r r ˆ s id π μ B d × = r r (29-1) • Symbol μ 0 is a constant called permeability constant: m/A T 10 26 1 m/A T 10 4 6 7 0 × × = - - . π μ • Equation (29-1) is known as the law of Biot and Savart . The law, which is deduced experimentally, is an inversed-square law Here is the unit vector that points from ds to P and r – the distance to P r ˆ
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• We have to first consider differential current-length element id s . The magnitude of the magnetic field d B created by this element at point P is given by Biot- Savart law: 2 0 sin θ ids μ dB = Magnetic Field Due to a Current in a Long Straight Wire • Lets derive equation for the magnetic field at perpendicular distance R from an infinitely long straight wire carrying current i 4 4 r π • The magnitude of the field produced by the lower half of the wire is exactly the same as that produced by the upper half. Therefore we can write: - = = = 0 2 0 0 sin 2 2 r θ ds π i μ dB dB B Here and 2 2 R s r + = 2 2 ) sin( sin R s R θ π θ + = - = • The direction of d B at point P is into the page. All other possible current-length elements of the wire have this same direction of d B at point P
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• So we get the integral: ( 29 π R i μ R s s π R i μ R s Rds π i μ B / / 2 ) ( 2 2 0 0 2 1 2 2 0 0 2 3 2 2 0 = + = + = i μ • Thus the magnitude of the magnetic field at a perpendicular distance R from a long straight wire carrying current i is: Magnetic Field Due to a Current in a Long Straight Wire 5 π R B 2 0 = • The direction of the magnetic field set up by a current-length element
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This document was uploaded on 11/04/2011 for the course PHY 108 at SUNY Buffalo.

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HR29_post - Chapter 29 Magnetic Fields due to Currents Will...

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