Biot-Savart Law The analogue of Coulomb’s Law is the Biot-Savart Law Consider a current loop ( I ) For element dℓ there is an associated element field dB dB perpendicular to both dℓ and r - r’ Inverse square dependence on distance  o /4  = Hm -1 Integrate to get Biot-Savart Law O r r’ r-r’ dB(r)dB(r)  dℓdℓ

Biot-Savart Law examples (1) Infinite straight conductor d ℓ and r, r’ in the page dB is into the page B forms concentric circles about the conductor B r - r’ dBdB I dℓdℓ r’ O r   z

Biot-Savart Law examples (2) Axial field of circular loop Loop perpendicular to page, radius a d ℓ out of page at top and r, r’ in the page On-axis element dB is in the page, perpendicular to r - r’, at  to axis. Magnitude of element dB Integrating around loop, only z-component of dB contributes net result r - r’dB || n I dℓdℓ z  dBzdBz r’ r a 

On-axis field of circular loop Introduce axial distance z, where |r-r’| 2 = a 2 + z 2 2 limiting cases r - r’dBdB I dℓdℓ z  dBzdBz r’ r a

Magnetic dipole moment The off-axis field of circular loop is much more complex. For z >> a (only) it is identical to that of the electric point dipole m = current times area vs p = charge times distance m = magnetic dipole moment r  m I q+ q- p

Current density Suppose current is composed of point charges Integrate current over some volume, V, say A d where A is cross-section area of wire d is element of length of wire n is a unit vector parallel to the current direction Current is commonly treated as continuous (j(r)), but is actually composed of point particles Current I and current density j

Continuity Equation Rate of charge entering xz face at y = 0: j y=0  x  z Cm -2 s -1 m 2 = Cs -1 Rate of charge leaving xz face at y =  y: j y=  y  x  z = (j y=0 + ∂j y /∂y  y)  x  z Net rate of charge entering cube via xz faces: (j y=0 - j y=  y )  x  z = -∂j y /∂y  x  y  z xx zz yy z y x Rate of charge entering cube via all faces: - (∂j x /∂x + ∂j y /∂y + ∂j z /∂z)  x  y  z = dQ encl /dt  = lim (  x  y  z)→0 Q encl /(  x  y  z) . j + d  /dt = 0 j y=  y j y=0 For steady currents d  /dt = 0 (Magnetostatics) and . j = 0 Applies to other j’s (heat, fluid, etc) Conservation of charge, energy, mass,..

Ampere’s Law Replace I d by j(r’) dr’ in Biot-Savart law O r r’ r-r’ dB(r)dB(r)  dℓdℓ See Homework Problems II for intermediate steps

Ampere’s Law Evaluate Div and Curl of B(r) NB ∇ acts on functions of r only, ∇ ’ acts on functions of r’ Absence of magnetic monopoles (generally valid) Ampere’s Law (limited to magnetostatics ( ∇.j = 0))

Ampere’s Law Can B(r) be expressed in terms of a potential? Yes! A is the vector potential

Integral form of law: enclosed current is integral dS of current density j Apply Stokes’ theorem Integration surface is arbitrary Must be true point wise Differential form of Ampere’s Law S j B dℓdℓ

B.dℓ for current loop Consider line integral B.dℓ from current loop of radius a Contour C is closed by large semi-circle, contributes zero to line integral   o I/2 oIoI z→+∞ z→-∞ I (enclosed by C) C a

Electric Magnetic Field reverses No reversal E.dℓ for electric dipole Consider line integral E.dℓ for electric dipole with charges ±q at ±a/2 Contour C is closed by large semi-circle, contributes zero to line integral

Ampere’s Law examples (1)Infinitely long, thin conductor B is constant on circle of radius r Exercise: Find radial profile of B inside conductor of radius R B B r R

(2) Solenoid with N loops/metre B constant and axial inside, zero outside Rectangular path, axial length L Exercise: Find B inside toroidal solenoid, i.e. one which forms a doughnut solenoid is to magnetostatics what capacitor is to electrostatics Ampere’s Law examples B I L