11/14/2004 section 7_3 The Biot-Savart Law blank.doc 1/1 7-3 The Biot-Savart Law and the Magnetic Vector Potential Reading Assignment: pp. 208-218 Q: Given some field B (r ) , how can we determine the source J ()r that created it? A: Easy! Æ JB()rxr= ∇ ( ) µ0 Q: OK, given some source J (r ), how can we determine what field B ()r it creates? A: HO: The Magnetic Vector Potential HO: Solutions to Ampere’s Law HO: The Biot-Savart Law Example: The Uniform, Infinite Line of Current HO: B-field from an Infinite Current Sheet Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Magnetic Vector Potential.doc 1/5 The Magnetic Vector Potential From the magnetic form of Gauss’s Law ∇ ⋅=B ()r0, it is evident that the magnetic flux density B (r ) is a solenoidal vector field. Recall that a solenoidal field is the curl of some other vector field, e.g.,: BA(rxr) = ∇ ( ) Q: The magnetic flux density B (r ) is the curl of what vector field ?? A: The magnetic vector potential A (r )! The curl of the magnetic vector potential A (r ) is equal to the magnetic flux density B (r ): ∇xrAB( ) = ( r) where: Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Magnetic Vector Potential.doc 2/5 ⎡Webers⎤ magnetic vector potential A ()r ⎣⎢ meter ⎦⎥ Vector field A ()r is called the magnetic vector potential because of its analogous function to the electric scalar potential V ()r . An electric field can be determined by taking the gradient of the electric potential, just as the magnetic flux density can be determined by taking the curl of the magnetic potential: EBA()r=−∇V ( r) (r) =∇ x() r Yikes! We have a big problem! There are actually (infinitely) many vector fields A ()r whose curl will equal an arbitrary magnetic flux density B ()r . In other words, given some vector field B (r ), the solution A (r ) to the differential equation ∇xrAB( ) = ( r) is not unique ! But of course, we knew this! To completely (i.e., uniquely) specify a vector field, we need to specify both its divergence and its curl. Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Magnetic Vector Potential.doc 3/5 Well, we know the curl of the magnetic vector potential A (r ) is equal to magnetic flux density B (r ). But, what is the divergence of A ()r equal to ? I.E.,: ∇ ⋅=A (r) ??? By answering this question, we are essentially defining A (r ). Let’s define it in so that it makes our computations easier! To accomplish this, we first start by writing Ampere’s Law in terms of magnetic vector potential: ∇=∇∇=xrBAJ() xxr( ) µ0 ( r) We recall from section 2-6 that: ∇∇xxAAA( r) =∇∇⋅( ( r)) −∇2 ( r) Thus, we can simplify this statement if we decide that the divergence of the magnetic vector potential is equal to zero: ∇ ⋅=A (r0) We call this the gauge equation for magnetic vector potential. Note the magnetic vector potential A (r ) is therefore also a solenoidal vector field. Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Magnetic Vector Potential.doc 4/5 As a result of this gauge equation, we find: ∇∇xxA( r) =∇(∇ ⋅A(r )) −∇2A( r) =−∇2A()r And thus Ampere’s Law becomes: 2 ∇=−∇=xrBAJ() ( r) µ0 ( r) Note the Laplacian operator ∇2 is the vector Laplacian, as it operates on vector field A (r ). Summarizing, we find the magnetostatic equations in terms of magnetic vector potential A (r ) are: ∇=xrAB( ) ( r) 2 ∇=−AJ()rrµ0 () ∇⋅A()r0 = Note that the magnetic form of Gauss’s equation results in the equation ∇⋅∇xrA () = 0. Why don’t we include this equation in the above list? Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Magnetic Vector Potential.doc 5/5 Compare the magnetostatic equations using the magnetic vector potential A ()r to the electrostatic equations using the electric scalar potential V ()r : E(r ) =−∇Vr( ) ρ ()r ∇⋅E()r = v ε0 Hopefully, you see that the two potentials A (r ) and V ()r are in many ways analogous. For example, we know that we can determine a static field E(r ) created by sources ρv ()r either directly (from Coulomb’s Law), or indirectly by first finding potential V (r ) and then taking its derivative (i.e., E()r =−∇Vr( )). Likewise, the magnetostatic equations above say that we can determine a static field B(r ) created by sources J()r either directly, or indirectly by first finding potential A ()r and then taking its derivative (i.e.,∇xrAB( ) = ( r)). ρv ()rVrr⇒⇒( ) E ( ) JAB()rrr⇒⇒() () Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 Solutions to Amperes Law.doc 1/4 Solutions to Ampere’s Law Say we know the current distribution J (r ) occurring in some physical problem, and we wish to find the resulting magnetic flux density B ()r . Q: How do we find B (r ) given J (r )? A: Two ways! We either directly solve the differential equation: ∇xrBJ( ) = µ0 ( r) Or we first solve this differential equation for vector field A ()r : 2 −∇AJ(rr) = µ0 ( ) and then find B ()r by taking the curl of A ()r (i.e., ∇=xrAB() () r). It turns out that the second option is often the easiest! To see why, consider the vector Laplacian operator if vector field A ()r is expressed using Cartesian base vectors: 22 2 2 ∇=∇+∇+∇A ()rrAaxx( ) ˆˆˆ Aa yy( r) Aa zz() r Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 Solutions to Amperes Law.doc 2/4 We therefore write Ampere’s Law in terms of three separate scalar differential equations: 2 ∇=−AJxx(rr) µ0 ( ) 2 ∇=−AJyy()rrµ0 () 2 ∇=−AJzz()rrµ0 () Each of these differential equations is easily solved. In fact, we already know their solution! Recall we had the exact same differential equation in electrostatcs (i.e., Poisson’s equation): −ρ (r ) ∇=2V ()r v ε0 We know the solution V (r ) to this differential equation is: 1 ρ (r′) V ()r = v dv ′ ∫∫∫ ′ 4rrπ ε0 V − Mathematically, Poisson’s equation is exactly the same as each of the three scalar differential equations at the top of the page, with these substitutions: 1 V ()r→→→Axv() r ρ ()r Jx () r µ0 ε0 Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 Solutions to Amperes Law.doc 3/4 The solutions to the magnetic differential equation are therefore: µ J (r′) Adv()r = 0 x ′ x ∫∫∫ ′ 4rrπ V − µ J ()r′ Adv()r = 0 y ′ y ∫∫∫ ′ 4rrπ V − µ J ()r′ Adv()r = 0 z ′ z ∫∫∫ ′ 4rrπ V − and since: ˆˆˆ A()rr=++AaAaAaxxyyzz( ) ( r) ( r) and: ˆˆˆ J()rr=++JaJaJaxxyyzz( ) ( r) ( r) we can combine these three solutions and get the vector solution to our vector differential equation: µ J (r′) A ()r = 0 dv ′ ∫∫∫ ′ 4rrπ V − Therefore, given current distribution J (r ), we use the above equation to determine magnetic vector potential A ()r . We then take the curl of this result to determine magnetic flux density B ()r . Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 Solutions to Amperes Law.doc 4/4 For surface current, the resulting magnetic vector potential is: µ J (r′) A ()r = 0 s ds ′ ∫∫ ′ 4rrπ S − and for a current I flowing along contour C, we find: µ IdA′ A ()r = 0 v∫ ′ 4rrπ C − Again, ponder the analogy between these equations involving sources and potentials and the equivalent equation from electrostatics: 1 ρ (r′) V ()r = v dv ′ ∫∫∫ ′ 4rrπ ε0 V − Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Biot Savart Law.doc 1/4 The Biot-Savart Law So, we now know that given some current density, we can find the resulting magnetic vector potential A (r ): µ J (r′) A ()r = 0 dv ′ ∫∫∫ ′ 4rrπ V − and then determine the resulting magnetic flux density B (r ) by taking the curl: BA(rxr) = ∇ ( ) Q: Golly, can’t we somehow combine the curl operation and the magnetic vector potential integral? A: Yes! The result is known as the Biot-Savart Law. Combining the two above equations, we get: µ J (r′) B ()rx=∇ 0 dv ′ ∫∫∫ ′ 4rrπ V − This result is of course not very helpful, but we note that we can move the curl operation into the integrand: Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Biot Savart Law.doc 2/4 µ J (r′) B ()rx=∇0 dv ′ ∫∫∫ ′ 4rrπ V − Note this result reverses the process: first we perform the curl, and then we integrate. We can do this is because the integral is over the primed coordinates (i.e.,r′) that specify the sources (current density), while the curl take the derivatives of the unprimed coordinates (i.e., r ) that describe the fields (magnetic flux density). Q: Yikes! That curl operation still looks particularly difficult. How we perform it? A: We take advantage of a know vector identity! The curl of vector field f ()rrG () , where f (r ) is any scalar field and G (r ) is any vector field, can be evaluated as: ∇=∇−∇xfr()()GGG() rff( r) x( r) ( rx) () r Note the integrand of the above equation is in the form ∇xfr()()G () r , where: 1 f ()r== and GJ()r() r′ rr− ′ Therefore we find: Jim Stiles The Univ. of Kansas Dept. of EECS 11/14/2004 The Biot Savart Law.doc 3/4 ⎛⎞J ()r′ 11 ⎛⎞ ∇=∇−∇xxrrx⎜⎟JJ()′′() ⎜⎟ ⎝⎠rr−−′′ rr ⎝⎠ rr − ′ In the first term we take the curl of J (r′).