PHY2049_06-08-11 - 1 Chapter 29 Magnetic Fields Due to...

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Unformatted text preview: 6/7/2011 1 Chapter 29 Magnetic Fields Due to Currents In this chapter we will explore the relationship between an electric current and the magnetic field it generates in the space around it. We will follow a two-prong approach, depending on the symmetry of the problem. For problems with low symmetry we will use the law of Biot-Savart in combination with the principle of superposition. For problems with high symmetry we will introduce Ampere’s law . Both approaches will be used to explore the magnetic field generated by currents in a variety of geometries (straight wire, wire loop, solenoid coil, toroid 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 (the Ampere) (29 – 1) A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, v proton /v alpha , is: A. 0.5 B. 1 C. 2 D. 4 E. 8 A proton (charge e), traveling perpendicular to a magnetic field in a circular orbit, experiences the same force as an alpha particle (charge 2e, mass 4p) which is also traveling perpendicular to the same field. The ratio of their orbital radii r proton /r alpha , is: A. 0.5 B. 1 C. 2 D. 4 E. 8 A proton (charge e), traveling perpendicular to a magnetic field in a circular orbit, experiences the same force as an alpha particle (charge 2e, mass 4p) which is also traveling perpendicular to the same field. The ratio of their orbital frequency f proton /f alpha is: This law gives the magnetic field generated by a wire segment of length that carries a current . Consider the geometry shown in the figure. Associated with the element dB ds i ds The law of Biot - Savart r we define an associated vector that has magnitude whch is equal to the length . The direction of is the same as that of the current that flows through segment . ds ds ds ds r r A 3 The magnetic field generated at point P by the element located at point A is given by the equation: . Here is the vector that connects 4 point A (location of element ) w o dB dS i ds r dB r r ds μ π × = r r r r r r 7 6 2 ith point P at which we want to determine . The constant 4 10 T m/A 1.26 10 T m/A and is known as sin " ". The magnitude of is: 4 Here is the angle o o dB i ds dB dB r μ π μ θ π θ-- = × ⋅ = × ⋅ = r r permeability constant between and . ds r r r 3 4 o i ds r dB r μ π × = r r r (29 – 2) The magnitude of the magnetic field generated by the wire at point P located at a distance from the wire is given by the equation: 2 o i B R R μ π = Magnetic field generated by a long straight wire 2 o i B R μ π = The magnetic field lines form circles that have their centers at the wire. The magnetic field vector is tangent to the magnetic field lines. The sense for is given by the . We po B B right hand rule r r int the thumb of the right hand in the...
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This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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PHY2049_06-08-11 - 1 Chapter 29 Magnetic Fields Due to...

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