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Lecture 9 Magnetoquasistatics In this lecture you will learn: • Basic Equations of Magnetoquasistatics • The Vector Potential • The Vector Poisson’s Equation • The Biot-Savart Law • Magnetic Field of Some Simple Current Carrying Elements • The Magnetic Current Dipole ECE 303 – Fall 2007 – Farhan Rana – Cornell University Equations of Magnetoquasistatics Equations of Electroquasistatics Equations of Magnetoquasistatics r r ∇. ε o E = ρ (r , t ) r ∇×E = 0 r r r ∂ε E ∇×H = J + o ∂t r ∇ . µo H = 0 r r r ∇ × H = J (r , t ) r r ∂ µoH ∇×E = − ∂t • Electric fields are produced by only electric charges • Magnetic fields are produced by only electric currents • Once the electric field is determined, the magnetic field can be found by the last equation above • Once the magnetic field is determined, the electric field can be found by the last equation above • Currents in magnetoquasistatics are solenoidal (i.e. with zero divergence) ( ) r r r ∇ . J (r , t ) = ∇ . ∇ × H = 0 In magnetoquasistatics the source of the magnetic field is electrical current ECE 303 – Fall 2007 – Farhan Rana – Cornell University 1 Ampere’s Law for Magnetoquasistatics r r r r ∫ H . ds = ∫∫ J . da r r ∇×H = J electric current density A closed contour Ampere’s Law: The line integral of magnetic field over a closed contour is equal to the total current flowing through that contour Right Hand Rule: The positive directions for the surface normal vector and of the contour are related by the right hand rule ECE 303 – Fall 2007 – Farhan Rana – Cornell University Magnetic Field of an Infinite Line-Current Consider an infinitely long line-current carrying a total current I in the +z-direction, as shown below Use ampere’s law on the closed contour shown by the dashed line: line current r r r r ∫ H . ds = ∫∫ J . da ⇒ (2π r ) Hφ (r ) = I ⇒ Hφ (r ) = I 2π r y r ds r x Working in the cylindrical coordinates Magnetic field is entirely in the φˆ direction and falls off as ~1/r from the line-current ECE 303 – Fall 2007 – Farhan Rana – Cornell University 2 Magnetic Field of a Solenoid H Consider a solenoid with N turns per unit length and carrying a current I Assumptions: • The magnetic field inside the solenoid is uniform and strong I • There is a fringing field outside the solenoid which is very weak and may be neglected I Hy r r r r ∫ H . ds = ∫∫ J . da ⇒ L H y = (LN ) I y L ⇒ H y = NI x ECE 303 – Fall 2007 – Farhan Rana – Cornell University The Vector Potential - I A Vector Identity: r For any vector F the divergence of the curl is always zero: ( ) r ∇. ∇×F = 0 The Vector Potential: In magnetoquasistatics the divergence of the B-field is always zero: ( ) ( ) r r ∇ . B = ∇ . µo H = 0 So one may represent the B-field as the curl of another vector: r r r B = µo H = ∇ × A r A is called the vector potential ⇒ ( ) r r r ∇ . B = ∇ . µo H = ∇ . ∇ × A = 0 ECE 303 – Fall 2007 – Farhan Rana – Cornell University 3 The Vector Potential - II In electroquasistatics we had: r ∇×E = 0 r E = −∇ φ Therefore we could represent the E-field by the scalar potential: ( ) ( ) r r ∇ . B = ∇ . µo H = 0 In magnetoquasistatics we have: Therefore we can represent the B-field by the vector potential: r r r B = µo H = ∇ × A A vector field can be specified (up to a constant) by specifying its curl and its divergence r Our definition of the vector potential A is not yet unique – we have only specified its curl r For simplicity we fix the divergence of the vector potential A to be zero: r ∇.A = 0 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Differential Equation for The Vector Potential Start from Ampere’s law in differential form: r r r ∇×H = J r Use: µo H = ∇ × A ( A Vector Identity: ( ) r r ∇ × ∇ × A = µo J To get: ( ) ( ) r r r ∇ × ∇ × F = ∇ ∇ . F − ∇ 2F r ) r r r ∇ ∇ . A − ∇ 2 A = µo J ⇒ r r ∇ 2 A = − µo J Use: ∇ . A = 0 The differential equation for the vector potential (also called the Vector Poisson’s Equation) r This is in fact 3 different equations (one for each component of A ) ∇ 2 Ax = − µo J x ∇ 2 Ay = − µo J y ∇ 2 Az = − µo J z ECE 303 – Fall 2007 – Farhan Rana – Cornell University 4 The Superposition Principle for The Vector Potential r r • Supposerforr the current distribution J1(r ) we have found the vector potential A1(r ) r r • Supposerforrsome other current distribution J 2 (r ) we have also found the vector potential A2 (r ) ( r r r r ) • Then the vector r rpotential r r A1(r ) + A2 (r ) is the solution for the current distribution J1(r ) + J 2 (r ) ( ) Proof: r r r r ∇ 2 A1(r ) = − µo J1(r ) + r r r r ∇ 2 A2 (r ) = − µo J 2 (r ) = ( ) ( ) r r r r r r r r ∇ A1(r ) + A2 (r ) = − µo J1(r ) + J 2 (r ) 2 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Recall the Superposition Integral for the Potential In the most general scenario, one has to solve the Poisson equation: r r ρ (r ) ∇ 2 φ (r ) = − r r r − r' εo r r' We know that the solution for a point charge sitting at the origin: r φ (r ) = r r q 4π ε o r r ρ (r ') To find the potential at any point one can sum up the contributions from different portions of a charge distribution treating each as a point charge r φ (r ) = ∫∫∫ r ρ (r ') r r dV ' 4π ε o r − r ' dV ' = dx ' dy ' dz ' A formal solution of the vector differential equation is the vector superposition integral: r r ∇ 2 A = − µo J r r r r µo J (r ') A(r ) = ∫∫∫ r r dV ' 4π r − r ' ECE 303 – Fall 2007 – Farhan Rana – Cornell University 5 The Biot-Savart Law - I Start from the superposition integral for the vector potential: r r r r µo J (r ') A(r ) = ∫∫∫ r r dV ' 4π r − r ' Now find the magnetic field: r r ∇×A H= µo r r r r ⎡ ⎤ J (r ' ) H (r ) = ∇ × ⎢ ∫∫∫ r r dV '⎥ 4 r − r ' π ⎣ ⎦ r r ⎛ 1 ⎞ r r 1 ∇ ⎜⎜ r r ⎟⎟ × J (r ') dV ' H (r ) = ∫∫∫ 4π ⎝ r − r' ⎠ r r r r J (r ') × nˆ rr '→ rr H (r ) = ∫∫∫ r r 2 dV ' 4π r − r ' A vector Identity: ( ) ( ) r r r ∇ × ϕ F = ϕ ∇ × F + (∇ϕ ) × F rˆ ⎛ 1⎞ r ⎝ ⎠ r r r ( 1 ⎞ r − r ') nr r r r ⎟ = − r r 3 = − r r '→r r2 r − r ' ⎟⎠ r − r' r − r' Recall that: ∇ ⎜ ⎟ = − r 2 ⎛ ∇ ⎜⎜ ⎝ Biot-Savart Law r nˆ rr '→ rr is a unit vector directed from point r ' on the current source r to the observation point r ECE 303 – Fall 2007 – Farhan Rana – Cornell University The Biot-Savart Law - II nˆ rr '→ rr Biot-Savart Law: r r r r J (r ') × nˆ rr '→ rr H (r ) = ∫∫∫ r r 2 dV ' 4π r − r ' r r r − r' P r r r r' r r J (r ') r nˆ rr '→ rr is a unit vector directed from point r ' on the current source r to the observation point r ECE 303 – Fall 2007 – Farhan Rana – Cornell University 6 Biot-Savart Law for Line-Currents Need to find a formula that gives the total magnetic field at a point due to a current r carrying wire as a superposition of magnetic ds ' field contributions from all the small pieces I of the wire y nˆ sr '→ rr r r r s' x z line-current carrying a current I Start from the Biot-Savart law: r r r r J (r ' ) × nˆ rr '→ rr H (r ) = ∫∫∫ r r 2 dV ' 4π r − r ' r r r r I (r ') × nˆ rr '→ rr J (r ') × nˆ rr '→ rr = ∫∫∫ r r 2 da' ds ' = ∫∫∫ r r 2 ds ' 4π r − r ' 4π r − r ' r I ds ' × nˆ sr '→ rr = ∫ r r2 4π r − s' Integrate over the cross-section area of the wire to get the total current carried by the wire ECE 303 – Fall 2007 – Farhan Rana – Cornell University Magnetic Field of a Finite Line-Current y Consider a line-current of length L with current I in the +x-direction. nˆ sr '→ rr Find the magnetic field at the point P on −L 2 the y-axis (as shown) dx ' Use the Biot-Savart law: z r r I ds ' × nˆ sr '→ rr H (0, y ,0 ) = ∫ r r2 4π r − s' P y x' r ds ' = xˆ dx ' I r ds '×nˆ sr '→ rr = zˆ L2 x y dx ' x '2 + y 2 r r 2 r − s ' = x '2 + y 2 = zˆ I 4π L2 ∫ −L 2 y dx ' (x '2 + y 2 )3 2 r As L → ∞, H (0, y ,0 ) → zˆ = zˆ ⎡ ⎤ L2 ⎥ ⎢ 2π y ⎢ y 2 + (L 2 )2 ⎥ ⎣ ⎦ I 2π y I …..recover the previous result ECE 303 – Fall 2007 – Farhan Rana – Cornell University 7 Magnetic Field of a Current Loop – Near Field y Consider a line-current in the form of a circular loop of radius a and carrying a current I, as shown in the figure r ds ' I P Find the magnetic field at the point P in the center of the loop x a Use the Biot-Savart law: r r I ds ' × nˆ sr '→ rr H (0,0,0 ) = ∫ r r 2 4π r − s' = zˆ r ds ' = φˆ' a dφ ' r ds '×nˆ sr '→ rr = zˆ a dφ ' r r 2 r − s' = a2 I 2 π a dφ ' I = zˆ ∫ 4π 0 a 2 2a ECE 303 – Fall 2007 – Farhan Rana – Cornell University Magnetic Field of a Current Loop – Far Field (Magnetic Dipole) P (r , θ , φ = 0 ) Consider a line-current in the form of a circular loop of radius a and carrying a current I, as shown in the figure (r >> a) Find the magnetic field at the point P far far away from the loop A small current loop such as this is a magnetic dipole z r θ r ds ' a y φ' x I Use the superposition integral for the A-field: r r r r ds ' µ J (r ' ) µ I A(r ,θ , φ = 0 ) = ∫∫∫ o r r dV ' = o ∫ r r 4π r − s ' 4π r − s ' r ds ' = φˆ' a dφ ' φˆ' = cos(φ ') yˆ − sin(φ ') xˆ r r r − s ' ≈ r − a sin(θ ) cos(φ ' ) Integrate over the cross-section of the wire Be careful – tricky integral – the unit vector is changing directions within the integral ECE 303 – Fall 2007 – Farhan Rana – Cornell University 8 Magnetic Field of a Current Loop – Far Field (Magnetic Dipole) r r r r µ J (r ' ) µ I ds ' A(r ,θ , φ = 0 ) = ∫∫∫ o r r dV ' = o ∫ r r 4π r − s ' 4π r − s ' P (r , θ , φ = 0 ) z µo I 2π [cos(φ ')yˆ − sin(φ ')xˆ ] a dφ ' ≈ ∫ r − a sin(θ ) cos(φ ') 4π 0 ≈ r y θ ( ) µo I π a 2 sin(θ ) yˆ 4π r2 φ' r ds ' a x I More generally: ( ) z r µo I π a 2 sin(θ ) φˆ A(r ,θ , φ ) ≈ 4π r2 And: r r ∇×A H (r , θ , φ ) = µo ≈ ( )[2 cos(θ )rˆ + sin(θ )θˆ ] 4π µ r 3 µo I π a 2 o ECE 303 – Fall 2007 – Farhan Rana – Cornell University Electric and Magnetic Dipoles and Dipole Moments z z H E +q I −q r r r ∇×A H ( r >> a ) = r r r E ( r >> d ) = −∇ φ (r ) ≈ qd 4πε o r 3 µo ( 2 cos(θ ) rˆ + sin(θ )θˆ ) ≈ ( )[2 cos(θ )rˆ + sin(θ )θˆ] 4π µ r 3 µo I π a 2 o Electric dipole moment: Magnetic dipole moment: ( ) r r p = qd r m = I π a 2 nˆ ECE 303 – Fall 2007 – Farhan Rana – Cornell University 9 Magnetic Flux and Vector Potential Line Integral The magnetic flux λ through a surface is the surface integral of the B-field through the surface r r B-field λ = ∫∫ B . da r r = µo ∫∫ H . da Since: r r r B = µo H = ∇ × A We get: r r λ = ∫∫ B . da ( ) r r = ∫∫ ∇ × A . da r r = ∫ A . ds r ds Stoke’s Theorem A closed contour The magnetic flux through a surface is equal to the line-integral of the vector potential along a closed contour bounding that surface ECE 303 – Fall 2007 – Farhan Rana – Cornell University ECE 303 – Fall 2007 – Farhan Rana – Cornell University 10