11/21/2004 section 7_4 Field Calculations using Amperes Law blank 1/1 - PDF document

11/21/2004 section 7_4 Field Calculations using Amperes Law blank 1/1 Jim Stiles The Univ. of Kansas Dept. of EECS 7-4 Field Calculations Using Ampere’s Law Q: Using the Biot-Savart Law is even more difficult than using Coloumb’s law. Is there an easier way? HO: B-field from Cylindrically Symmeteric Current Distributions Example: A Hollow Tube of Current Example: The B-field of a Coaxial Transmission Line HO: Solenoids 11/21/2004 B-Field from Cylindrically Symmetric Current Distributions 1/4 Jim Stiles The Univ. of Kansas Dept. of EECS B-Field from Cylindrically Symmetric Current Distributions in Section 4-5. We found that distribution is a function of coordinate r Similarly, we can define a cylindrically symmetric currentdistribution. A current density and is a function of rˆ independent and produce a magnetic flux density of the form: rBa 11/21/2004 B-Field from Cylindrically Symmetric Current Distributions 3/4 Jim Stiles The Univ. of Kansas Dept. of EECS This is a special contour, called the Amperian Path for cylindrically symmetric current densities. To see why it is , let us use it in the cylindrically symmetric form of BadIBaadφφφρρφρρφπρρµ From this result, we can conclude that: enc But what is We of course can determine this by integrating the J across the surface of this circular aperture ( dsadd 11/21/2004 Example A Hollow Tube of Current 1/7 Jim Stiles The Univ. of Kansas Dept. of EECS Example: A Hollow Tube of Current Consider a hollow cylinder of uniform current, flowing in the direction: surface of the hollow cylinder has , while the ˆza 0ˆrzJ x 11/21/2004 Example A Hollow Tube of Current 4/7 Jim Stiles The Univ. of Kansas Dept. of EECS and therefore: 0 for B cylinder is Note for ˆrzJ (i.e., 0  zzzJdJdJddJd ρρρρρρρρρρρρ′′′′′′′′′′′′′∫∫∫ 11/21/2004 Example A Hollow Tube of Current 6/7 Jim Stiles The Univ. of Kansas Dept. of EECS Summarizing, we find that the produced by this hollow tube of current is: bWebersabc= expression by determining the total We of course can determine by performing the rJ across the cross sectional surface S of the cylinder: IdsJaaddJddJcb ρρφ 11/21/2004 Example The B-Field of a Coaxial Transmission Line 1/6 Jim Stiles The Univ. of Kansas Dept. of EECS Example: The B-Field of Coaxial Transmission Line Consider now a za I 0 b c surface of the conductor has radius surface of the conductor has radius radius of conductor has radius 11/21/2004 Example The B-Field of a Coaxial Transmission Line 2/6 Jim Stiles The Univ. of Kansas Dept. of EECS Typically, the flowing on the conductor is that flowing in the conductor. Thus, if is flowing in the conductor in the direction then current will be flowing in the conductor in the ) direction. Hey! If there is current, a magnetic flux density must We’ve already determined this (sort of) ! Recall we found the magnetic flux density produced by a hollow cylinder—we can use this to determine the magnetic flux density in a coaxial transmission line. find it necessary to point out that you are innernot Mathematically, we can view cylinder with an outer radius 11/21/2004 Example The B-Field of a Coaxial Transmission Line 3/6 Jim Stiles The Univ. of Kansas Dept. of EECS Thus, we can use the results of the previous handout to conclude that the magnetic flux density produced by the 0000222202inneraaaWebersπρπLikewise, we can use the same result to determine the magnetic flux density of the current flowing in the outer conductor: 222outerbWebersabccbm=Note the ) in the outer conductor. 11/21/2004 Example The B-Field of a Coaxial Transmission Line 5/6 Jim Stiles The Univ. of Kansas Dept. of EECS a b rrrouterinner BBB 0000rrrouterinner πρπρBBB 0000rrrouterinner πρπρ BBB 11/21/2004 Solenoids 2/3 Jim Stiles The Univ. of Kansas Dept. of EECS to find the magnetic flux density resulting from this structure. The result is: Jaa Note the direction of the magnetic flux density is in the direction --it points down independent of solenoid We can easily make a solenoid by forming a Yeah right! How are we supposed to get current to don’t see how this is even turns leakage flux lines I 11/21/2004 Solenoids 3/3 Jim Stiles The Univ. of Kansas Dept. of EECS approximately equal to: JNI is the number of turns/unit length. Inserting find the magnetic flux density NIa