Biot-Savart Law

1. I dB(r) r-r’ r O r’ dℓ 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/4p = 10-7 Hm-1 • Integrate to get Biot-Savart Law

2. I q dℓ r’ z r - r’ dB a O r B Biot-Savart Law examples (1) Infinite straight conductor dℓ and r, r’ in the page dB is into the page B forms concentric circles about the conductor

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

4. dℓ r - r’ dB r’ q r I z a dBz On-axis field of circular loop Introduce axial distance z, where |r-r’|2 = a2 + z2 2 limiting cases

5. p m q+ q- q r I 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 vsp = charge times distance m = magnetic dipole moment

6. Current I and current density j Current density Suppose current is composed of point charges Integrate current over some volume, V, say A d𝓁 where A is cross-section area of wire d𝓁 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

7. z Dz y Dy Dx x Continuity Equation Rate of charge entering xz face at y = 0: jy=0DxDz Cm-2s-1 m2 = Cs-1 Rate of charge leaving xz face at y = Dy: jy=DyDxDz = (jy=0 + ∂jy/∂yDy) DxDz Net rate of charge entering cube via xz faces: (jy=0 -jy=Dy )DxDz = -∂jy/∂yDxDyDz Rate of charge entering cube via all faces: -(∂jx/∂x + ∂jy/∂y + ∂jz/∂z)DxDyDz = dQencl/dt r= lim (DxDyDz)→0 Qencl/(DxDyDz) . j + dr /dt = 0 jy=0 jy=Dy For steady currents dr /dt = 0(Magnetostatics) and . j = 0 Applies to other j’s (heat, fluid, etc) Conservation of charge, energy, mass, ..

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

9. 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))

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

11. B j dℓ S Differential form of Ampere’s Law 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

12. I (enclosed by C) a z→-∞ z→+∞ C moI moI/2  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

13. 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

14. B B r R Ampere’s Law examples Infinitely long, thin conductor B is constant on circle of radius r Exercise: Find radial profile of Binside conductor of radius R

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