7-4 Field Calculations Using Ampere s Law

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1 11/21/24 section 7_4 Field Calculations using Amperes Law lank 1/1 7-4 Field Calculations Using Ampere s Law Q: Using the iot-savart Law is even more difficult than using Coloum s law. Is there an easier way? A: HO: -field from Cylindrically Symmeteric Current Distriutions Example: A Hollow Tue of Current Example: The -field of a Coaxial Transmission Line HO: Solenoids

2 11/21/24 -Field from Cylindrically Symmetric Current Distriutions 1/4 -Field from Cylindrically Symmetric Current Distriutions Recall we discussed cylindrically symmetric charge distriutions in Section 4-5. We found that a cylindrically symmetric charge distriution is a function of coordinate only r = ). (i.e., ( ) ( ) v v Similarly, we can define a cylindrically symmetric current distriution. A current density J ( r ) is said to e cylindrically symmetric if it points in the direction ˆa and is a function of coordinate only: J ( r ) = J ( ) ˆa In other words, J = J =, and J is independent of oth coordinates and. We find that a cylindrically symmetric current density will always produce a magnetic flux density of the form: ( ) = ( ) ˆ r a

3 11/21/24 -Field from Cylindrically Symmetric Current Distriutions 2/4 In other words, = =, and is independent of oth coordinates and. Now, lets apply these results to the integral form of Ampere s Law: r d = ˆa d = Ienc C ( ) ( ) where you will recall that I enc is the total current flowing through the aperture formed y contour C: C I enc C Say we choose for contour C a circle, centered around the - axis, with radius. d = ˆa d Amperian Path for Cylindrically Symmetric Current Distriutions C

4 11/21/24 -Field from Cylindrically Symmetric Current Distriutions 3/4 This is a special contour, called the Amperian Path for cylindrically symmetric current densities. To see why it is special, let us use it in the cylindrically symmetric form of Ampere s Law: ˆa d = I 2π C ( ) ( ) ˆa ˆa d = ( ) 2π d = ( ) 2π = I enc enc From this result, we can conclude that: ( ) = I enc 2π Q: ut what is I enc? A: The current flowing through the circular aperture formed y contour C! We of course can determine this y integrating the current J r across the surface of this circular aperture density ( ) ( ds = ˆa d d ):

5 11/21/24 -Field from Cylindrically Symmetric Current Distriutions 4/4 I enc S 2π ( r ) = J ds ( ) = J ˆa ˆa d d ( ) = 2π J d Comining these results, we find that the magnetic flux density r J r ( ) created y a cylindrically symmetric current density ( ) is: ( r ) = I enc 2π = ˆa ˆa J ( ) d For example, consider again a wire with current I flowing along the -axis. This is a cylindrically symmetric current, and the total current enclosed y an Amperian path is clearly I for all (i.e., I enc =I ). From the expression aove, the magnetic flux density ( r ) is therefore: ( r ) = I ˆ 2π a The same result as determined y the iot-savart Law!

6 11/21/24 Example A Hollow Tue of Current 1/7 Example: A Hollow Tue of Current Consider a hollow cylinder of uniform current, flowing in the a ˆ direction: J ( r ) ˆ = J a a ˆ y The inner surface of the hollow cylinder has radius, while the outer surface has radius c. c x

7 11/21/24 Example A Hollow Tue of Current 2/7 The current density in the hollow cylinder is uniform, thus we J r as: can express current density ( ) J < Amps m > c ( r) Jaˆ < < c 2 = Q: What magnetic flux density ( r ) is produced y this current density J ( r )? A: We could use the iot-savart Law to determine ( r ), ut note that J ( r ) is cylindrically symmetric! In other words, current density J ( r ) has the form: J ( r ) = J ( ) ˆ a The current is cylindrically symmetric! I suggest you use my law to determine the resulting magnetic flux density.

8 11/21/24 Example A Hollow Tue of Current 3/7 Recall using Ampere s Law, we determined that cylindrically symmetric current densities produce magnetic flux densities of the form: ( r ) = I enc 2π ˆa = ˆa J ( ) d Therefore, we must evaluate the integral for the current density in this case. ecause of the piecewise nature of the current density, we must evaluate the integral for three different cases: 1) when the radius of the Amperian path is less than (i.e., < ). 2) when the radius of the Amperian path is greater than ut less than c (i.e., < < c). < 3) when the radius of the Amperian path is greater than c. Note for <, J ( r) = and therefore the integral is ero: ( ) J d = d =

9 11/21/24 Example A Hollow Tue of Current 4/7 and therefore: r ( ) ˆ = a = for < Thus, the magnetic flux density in the hollow region of the cylinder is ero! < < c Note for < < c, J ( r ) = J ˆ a (i.e., J ( ) = J) and therefore: ( ) = ( ) + ( ) J d J d J d = d + J d = + J = J = J d 2

10 11/21/24 Example A Hollow Tue of Current 5/7 and therefore the magnetic flux density in the non-hollow portion of the cylinder is: > c ( r ) = ˆ a J for < < c 2 Note that outside the cylinder (i.e., J ( r ) is again ero, and therefore: > c ), the current density ( ) = ( ) + ( ) + ( ) J d J d J d J d c c c = d + J d + d = + J d + c c = J c 2 = J Thus, the magnetic flux density outside the current cylinder is: c c ( r ) = ˆ a J for c > 2

11 11/21/24 Example A Hollow Tue of Current 6/7 Summariing, we find that the magnetic flux density produced y this hollow tue of current is: ( r ) < J Weers ˆ = a < < c 2 2 m J c ˆ a 2 > c We can find an alternative expression y determining the total current flowing through this cylinder (let s call this current I ). We of course can determine I y performing the surface integral of the current density J ( r ) across the cross sectional surface S of the cylinder: I S 2π c ( r ) = J ds = J ˆ a ˆ a d d 2π c = J d d ( ) = J π c

12 11/21/24 Example A Hollow Tue of Current 7/7 Therefore, we can conclude that: J = I π ( c ) Inserting this into the expression for the magnetic flux density, we find: ( r ) < I Weers ˆ = a c < < 2 2π c m I 2π ˆ a > c Note the field outside of the cylinder ( > c ) ehaves precisely as would the field from a wire of current I! c

13 11/21/24 Example The -Field of a Coaxial Transmission Line 1/6 Example: The -Field of Coaxial Transmission Line Consider now a coaxial cale, with inner radius a : I I a ˆ The outer surface of the inner conductor has radius a, the inner surface of the outer conductor has radius, and the outer radius of the outer conductor has radius c. a c

14 11/21/24 Example The -Field of a Coaxial Transmission Line 2/6 Typically, the current flowing on the inner conductor is equal ut opposite that flowing in the outer conductor. Thus, if current I is flowing in the inner conductor in the direction a ˆ, then current I will e flowing in the outer conductor in the opposite (i.e., a ) direction. ˆ Q: Hey! If there is current, a magnetic flux density must r? e created. What is the vector field ( ) A: We ve already determined this (sort of)! Recall we found the magnetic flux density produced y a hollow cylinder we can use this to determine the magnetic flux density in a coaxial transmission line. A coaxial cale can e viewed as two hollow cylinders! Q: I find it necessary to point out that you are indeed wrong the inner conductor is not hollow! A: Mathematically, we can view the inner conductor as a hollow cylinder with an outer radius a and an inner radius of ero!

15 11/21/24 Example The -Field of a Coaxial Transmission Line 3/6 Thus, we can use the results of the previous handout to conclude that the magnetic flux density produced y the current flowing in the inner conductor is: inner ( r ) I I ˆ a ˆ a a = 2 < 2π a 2π a Weers m I ˆ a > a 2π = 2 Likewise, we can use the same result to determine the magnetic flux density of the current flowing in the outer conductor: outer ( r ) < I Weers ˆ = a c < < 2 2π c m I 2π ˆ a > c Note the minus sign is due to direction of the current ( a ) in the outer conductor. ˆ

16 11/21/24 Example The -Field of a Coaxial Transmission Line 4/6 We can now apply superposition to determine the total magnetic flux density in a coaxial transmission line! Specifically: if J( r) = J ( r) + J ( r) inner outer then ( r) = ( r) + ( r) inner outer Note due to the piecewise nature of these solutions, we must evaluate this sum for 4 distinct regions: 1) < a (in the inner conductor) 2) a < < (in the region etween the conductors) 3) < < c (in the outer conductor) 4) > c (outside the coaxial cale) < a ( r) = ( r) + ( r) inner outer I = ˆ 2 a + 2π a I = ˆ a 2 2π a

17 11/21/24 Example The -Field of a Coaxial Transmission Line 5/6 a < < < < c > c ( r) = ( r) + ( r) inner I = ˆ a + 2π I = ˆ a 2π ( r) = ( r) + ( r) outer inner outer ˆ a I I = + ˆ a 2π 2π c I c ˆ a 2π c = ( r) = ( r) + ( r) inner outer I I = ˆ a ˆ + a 2π 2π =

18 11/21/24 Example The -Field of a Coaxial Transmission Line 6/6 Summariing, we find the total magnetic flux density to e: ( r ) I ˆ 2 a < a 2π a I ˆ a a < < 2π Weers = 2 m I c ˆ a c < < 2π c > c ( ) a c The magnetic flux density and the electric field outside of a coaxial transmission line are ero!

19 11/21/24 Solenoids 1/3 Solenoids An important structure in electrical and computer engineering is the solenoid. A solenoid is a tue of current. However, it is different from the hollow cylinder example, in that the current flows around the tue, rather than down the tue: a J s ( r ) Aligning the center of the tue with the -axis, we can express the current density as: J s < a Amps r J ˆ = s a = a m > a ( ) where a is the radius of the solenoid, and J s is the surface current density in Amps/meter.

20 11/21/24 Solenoids 2/3 We can use Ampere s Law to find the magnetic flux density resulting from this structure. The result is: ( r ) J a < a = > a ˆ s Note the direction of the magnetic flux density is in the direction aˆ --it points down the center of the solenoid. Note also that the magnitude ( r ) is independent of solenoid radius a! Q: Yeah right! How are we supposed to get current to flow around this tue? I don t see how this is even possile. A: We can easily make a solenoid y forming a wire spiral around a cylinder. N turns L I leakage flux lines

21 11/21/24 Solenoids 3/3 The surface current density J s of this solenoid is approximately equal to: NI Js = = N I L where N = N L is the numer of turns/unit length. Inserting this result into our expression for magnetic flux density, we find the magnetic flux density inside a solenoid: NI ˆ a L = N I ˆ a ( r ) =

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