Phys2212_33.5+to+33.10

# Phys2212_33.5+to+33.10 - Physics 2212 Electricity and...

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Physics 2212 Electricity and Magnetism Lecture 22 (Knight: 33.5 to 33.10) Faraday’s Law

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10/06/09 Physics 2212 - Lecture 22 2 Question What is the ranking of the forces in the figure? (a) F 1 =F 2 =F 3 =F 4 ; (b) F 1 <F 2 =F 3 >F 4 ; (c) F 1 =F 3 <F 2 =F 4 ; (d) F 1 =F 4 <F 2 =F 3 ; (e) F 1 <F 2 <F 3 =F 4 ;
10/06/09 Physics 2212 - Lecture 22 3 Magnetic Flux The number of arrows passing through the loop depends on two factors: (1) The density of arrows, which is proportional to B (1) The effective area A eff = A cos θ of the loop We use these ideas to define the magnetic flux: m eff : cos Flux A B AB θ Φ ≡ = 2 : 1 weber = 1 Wb = 1 Tm Flux units

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10/06/09 Physics 2212 - Lecture 22 4 Area Vector m eff cos A B AB A B θ Φ ≡ = = r r Define the area vector A of a loop such that it has the loop area as its magnitude and is perpendicular to the plane of the loop. If a current is present, the area vector points in the direction given by the thumb of the right hand when the fingers curl in the direction of current flow. If the area is part of a closed surface, the area vector points outside the enclosed volume. With this definition:
10/06/09 Physics 2212 - Lecture 22 5 Example : A Circular Loop Rotating in a Magnetic Field The figure shows a 10 cm diameter loop rotating in a uniform 0.050 T magnetic field. What is the magnitude of the flux through the loop when the angle is θ =0 0 , 30 0 , 60 0 , and 90 0 ? 2 2 3 2 (0.005 m) 7.85 10 m A R π - = = = × 4 4 m 4 3.93 10 Wb for 0 3.40 10 Wb for 30 cos 1.96 10 Wb for 60 0 Wb for 90 AB θ - - - × = ° × = ° Φ = = × = ° = °

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10/06/09 Physics 2212 - Lecture 22 6 Magnetic Flux in a Nonuniform Field So far, we have assumed that the loop is in a uniform field. What if that is not the case? The solution is to break up the area into infinitesimal pieces, each so small that the field within it is essentially constant. Then: m d B dA Φ = r r m area of loop B dA Φ = r r
10/06/09 Physics 2212 - Lecture 22 7 Example : Magnetic Flux from a Long Straight Wire The near edge of a 1.0 cm x 4.0 cm rectangular loop is 1.0 cm from a long straight wire that carries a current of 1.0 A, as shown in the figure. What is the magnetic flux through the loop? dA bdx = 0 2 4 I B x μ π = 0 2 4 m dx d B dA Ib x Φ = = r r 0 0 0 2 2 ln 2 ln 4 4 4 c a c a m c c dx c a Ib Ib x Ib x c + + + Φ = = = 9 5.55 10 Wb m - Φ = ×

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10/06/09 Physics 2212 - Lecture 22 8 Lenz’s Law (1) Heinrich Friedrich Emil Lenz (1804-1865) In 1834, Heinrich Lenz announced a rule for determining the direction of an induced current, which has come to be known as Lenz’s Law. Here is the statement of Lenz’s Law: There is an induced current in a closed conducting loop if and only if the magnetic flux through the loop is changing. The direction of the induced current is such that the induced magnetic field opposes the change in the flux.
10/06/09 Physics 2212 - Lecture 22 9 Lenz’s Law (2)

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Phys2212_33.5+to+33.10 - Physics 2212 Electricity and...

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