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- length element turns out to be: The law of Biot-Savart 3 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 4 r θ ids π 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 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 B 0 = 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 2 The direction of the magnetic field set up by a current-length element id s is defined by a right-hand rule:
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