FI 2201 Electromagnetism Alexander A. Iskandar, Ph.D. Physics of Magnetism and Photonics Research Group Magnetostatics BIOT SAVART LAW AND AMPERE’S LAW 1 Steady Current • Stationary charges produced electric fields that is constant in time → Electrostatics • Current that does not change with time (steady current) produced magnetic fields that is constant in time → Magnetostatics • Since current does not change with time, this means that there is no built-up of charges, hence ∂ρ = 0 ∂t and by continuity equation r ∇ ⋅ J = 0 Alexander A. Iskandar Electromagnetism 3 Biot-Savart Law • Magnetic field of a steady line current is given by r r r µ I ׈r µ I dl′׈r r B()r = 0 dl′ = 0 I 4π ∫ r 2 4π ∫ r 2 r • The constant µ0 is called vacuum permeability dl′ −7 2 and it has a value of µ0 = 4π ×10 N A • The reason it can be taken to be exact is because of the relation it bears to the free-space permittivity and the 2 speed of light in vacuum c = 1 ε 0 µ 0 , where c = 2.99792458 × 108 m/s. • Nature only constrains the product of the free-space permeability and permittivity, so we can define one, and determine the other by experimentation. r • With this, the units of B is given by 1 T = 1 N/(A.m) Alexander A. Iskandar Electromagnetism 4 2 Biot-Savart Law • Note the similarity of Biot-Savart Law and Coulomb’s Law r r 1 λdl′ˆr r µ0 Idl′׈r E r = B(r ) = ( ) ∫ 2 ∫ 2 4πε0 r 4π r r r r 1 σda′ˆr r µ0 K(r′)׈r E r = B()r = da′ () ∫ 2 ∫ 2 4πε0 r 4π r r r r 1 ρdτ ′ˆr r µ J (r′)׈r E r = B()r = 0 dτ ′ () ∫ 2 ∫ 2 4πε0 r 4π r • Example 5.5 and 5.6 • Problem 5.9b Alexander A. Iskandar Electromagnetism 5 Divergence and Curl • Any vector field is uniquely specified by its divergence and curl. • Consider a volume V containing current I,,y current density r J ()r′ r r r µ J ()r′ × r r µ ⎛ J (r′)× r ⎞ B r = 0 dτ ′ → ∇ ⋅ B = 0 ∇ ⋅⎜ ⎟dτ ′ () ∫ 2 ∫ ⎜ 2 ⎟ 4π r 4π ⎝ r ⎠ • By product rule r ⎛ J ()r′ ׈r ⎞ ˆr r r ⎛ ˆr ⎞ ∇ ⋅⎜ ⎟ = ⋅ ∇× J r′ − J r′ ⋅ ∇× ⎜ 2 2 ()() ()⎜ 2 ⎟ ⎝ r ⎠ r ⎝ r ⎠ but r ⎛ ˆr ⎞ ∇× J ()r′ = 0 ⎜∇× ⎟ = 0 ⎝ r 2 ⎠ • Thus, r r r ∇⋅ B = 0 ∫ B ⋅da = 0 Alexander A. Iskandar Electromagnetism S 6 3 Divergence and Curl r • Taking the curl of B ( r ) yields r r µ ⎛ J (r′)× r ⎞ ∇× B = 0 ∇×⎜ ⎟dτ ′ ∫ ⎜ 2 ⎟ 4π ⎝ r ⎠ • By product rule r r r r r r r r r r ∇×(A× B)= (B ⋅∇)A − (A⋅∇)B + A(∇⋅ B)− B(∇⋅ A) r then since the current density is a function of r ′, J ( r ′ ) , all r derivative of J () r ′ gives zero, hence r ⎛ J ()r′ ׈r ⎞ r ⎛ ˆr ⎞ r ˆr ∇×⎜ ⎟ = J r′ ∇⋅ − J r′ ⋅∇ ⎜ 2 ()⎜ 2 ( () ) 2 ⎝ r ⎠ ⎝ r ⎠ r but ˆr 3 ∇⋅ = 4πδ ()r r 2 and r ˆr r ˆr − ()J ()r′ ⋅∇ = (J ()r′ ⋅∇′) r 2 r 2 Alexander A. Iskandar Electromagnetism 7 Divergence and Curl • Consider only the x-component and using the product rule r r r ∇ ⋅(fA)= (A⋅∇)f + f (∇ ⋅ A) we have r (x-x′) ⎡(x-x′) r ⎤ (x-x′) r J r′ ⋅∇′ = ∇′⋅ J r′ − ∇′⋅ J r′ ()() 3 ⎢ 3 ()⎥ 3 ( ()) r ⎣ r ⎦ r • But for steady current we have r ∇′⋅ J (r′) = 0 and ⎡()x-x′ r ⎤ (x-x′) r ∇′⋅ J r′ dτ ′ = J r′ ⋅dar′ ∫ ⎢ 3 ( )⎥ ∫ 3 ( ) V ⎣ r ⎦ S r Taking a very large volume, this surface integral is zero r since J () r ′ on the boundary of this volume is zero. • Thus r µ r r ∇× B = 0 J ()r′ 4πδ 3 (r − r′ )dτ ′ = µ J ()r 4π ∫ 0 Alexander A. Iskandar Electromagnetism 8 4 Divergence and Curl •Or, r r r ∇× B = µ0 J (r ) • And in integral form r r r r ∇× B ⋅dar′ = B ⋅dl = µ J ⋅dar′ = µ I ∫ ∫ 0 ∫ 0 enc S C S or r r B ⋅dl = µ I ∫ 0 enc C This is known as Ampere ’s Law. • Example 5.7 and 5.8 Alexander A. Iskandar Electromagnetism 9 Summary of Static Fields • Governing equation of electrostatics and magnetostatics fields are r ρ r ∇⋅ E = ∇ ⋅ B = 0 ε 0 r r r r ∇× E = 0 ∇× B = µ0 J (r ) r Q r r ∫ E ⋅dar = enc ∫ B ⋅da = 0 S ε 0 S r r r r E ⋅dl = 0 B ⋅dl = µ I ∫ ∫ 0 enc C C Alexander A. Iskandar Electromagnetism 10 5 More Examples Problem 5.13 • A steady current I flows down a long cylindrical wire of radius a. Find the magnetic field both inside and outside the wire, if The current is uniformly distributed over the outside surface of the wire The current is uniformly distributed in such a way that J is proportional to s, the distance from the axis. Alexander A. Iskandar Electromagnetism 11 More Examples Problem 5.13 Y • Use Ampere’s Law a K • For the case of only surface current Z s s r r B ⋅dl = µ0 Ienc → B(2πs) = 0 → B = 0 ∫ X C1 C1 r r µ I C2 B ⋅dl = µ I → B(2πs) = µ I → B = 0 ∫ 0 enc 0 2πs C2 Thus, ⎧0 s < a r ⎪ B = µ I ⎨ 0 ϕˆ s > a ⎩⎪ 2πs Alexander A. Iskandar Electromagnetism 12 6 More Examples Problem 5.13 Y • For the case of volume current which increases linearly with a J distance from the axis, the current density Z s s J = J 0 a ds X with J0 is determined as follows r a s 2πJ a3 2 I = J ⋅dar = J 2πsdds = 0 = πa2 J ∫ ∫ 0 ( ) 0 0 a a 3 3 Thus, 3I 3I J = → J = s 0 2πa2 2πa3 Alexander A. Iskandar Electromagnetism 13 More Examples Problem 5.13 Y • Then for C1, the Ienc r s 3Is′ Is3 I = J ⋅dar = 2πs′ds′ = a J enc ∫ ∫ 3 ( ) 3 0 2πa a Inserting to Ampere’s Law Z s s r r Is3 B ⋅dl = µ I → B()2πs = µ ∫ 0 enc 0 a3 X C1 C1 or C2 µ I B = 0 s2 2πa3 Thus, ⎧ µ I 0 s2ϕˆ s < a r ⎪ 3 B = 2πa ⎨ µ I ⎪ 0 ϕˆ s > a ⎩ 2πs Alexander A. Iskandar Electromagnetism 14 7 More Examples A long cylindrical wire with a hole Y • A cylindrical wire of radius 2R has a hole of radius R as depicted in R the figure. The wire carry constant current density. What is the magnetic field in the hole ? Z 2R X Alexander A. Iskandar Electromagnetism 15 More Examples A long cylindrical wire with a hole Y • This problem can be solved using Ampere’ s Law and superposition R principle. Z 2R X R = + x r 2R B r 2 B1 Alexander A. Iskandar Electromagnetism 16 8 More Examples A long cylindrical wire with a hole Y • The current density I I ϕˆ1 J = 2 2 = 2 π ()4R − R 3πR s2 R s • Then for the larger wire, the 1 ϕ r magnetic field at position s1, with X Ampere’s Law 2R 2 r µ Js µ Is B (2πs ) = µ Jπ (s ) B = 0 1 ϕˆ = 0 1 ϕˆ 1 1 0 1 1 2 1 6πR2 1 µ I x2 + y 2 = 0 ()− xˆ sinϕ + yˆ cosϕ 6πR2 µ I = 0 ()− yxˆ + xyˆ 6πR2 Alexander A. Iskandar Electromagnetism 17 More Examples A long cylindrical wire with a hole Y • For the smaller wire, the magnetic field at position sr from a shifted 2 s coordinate system is 2 R r µ0 Js2 µ0 Is2 X’ B′ = − ϕˆ = − ϕˆ x s1 2 2 2 6πR2 2 X µ I ()x′ 2 + ()y′ 2 = − 0 ()− ˆx sinϕ′ + ˆy cosϕ′ 2R 6πR2 where, x′ y′ cosϕ′ = , sinϕ′ = ()x′ 2 + ()y′ 2 ()x′ 2 + ()y′ 2 and x′ = x and y′ = y − R Alexander A. Iskandar Electromagnetism 18 9 More Examples A long cylindrical wire with a hole Y • Thus r µ I B = 0 (− [y − R]ˆx + xˆy) 2 6πR 2 • And the total field in the hollow region is X r r r B = B1 + B2 µ I µ I = 0 (− yˆx + xˆy)− 0 (− [y − R]ˆx + xˆy) 6πR2 6πR2 µ I = − 0 ˆx a constant !! 6πR Alexander A. Iskandar Electromagnetism 19 10.