4 Amperes Law and Applications As far as

4). Ampere’s Law and Applications • As far as possible, by analogy with Electrostatics • B is “magnetic flux density” or “magnetic induction” • Units: weber per square metre (Wb m-2) or tesla (T) • Magnetostatics in vacuum, then magnetic media based on “magnetic dipole moment”

Biot-Savart Law • • The analogue of Coulomb’s Law is the Biot-Savart Law d. B(r) r r-r’ Consider a current loop (I) O • I For element dℓ there is an associated element field d. B perpendicular to both dℓ’ and r-r’ same 1/(4 pr 2) dependence o is “permeability of free space” defined as 4 p x 10 -7 Wb A-1 m-1 Integrate to get B-S Law r’ dℓ’

B-S Law examples (1) Infinitely long straight conductor dℓ and r, r’ in the page d. B is out of the page B forms circles centred on the conductor Apply B-S Law to get: I dℓ q r’ z O r - r’ r a d. B q = p/2 + a sin q = cos a = B

B-S Law examples (2) “on-axis” field of circular loop dℓ Loop perpendicular to page, radius a dℓ out of page and r, r’ in the page On-axis element d. B is in the page, perpendicular to r - r’, at q to axis. r - r’ I r’ a r z Magnitude of element d. B Integrating around loop, only z-components of d. B survive The on-axis field is “axial” d. B q d. Bz

On-axis field of circular loop dℓ r - r’ I r’ a Introduce axial distance z, where |r-r’|2 = a 2 + z 2 2 limiting cases: r z d. B q d. Bz

Magnetic dipole moment The off-axis field of circular loop is much more complex. For z >> a it is identical to that of the electric dipole m m “current times area” vs p “charge times distance” q r

B field of large current loop • • Electrostatics – began with sheet of electric monopoles Magnetostatics – begin sheet of magnetic dipoles Sheet of magnetic dipoles equivalent to current loop Magnetic moment for one dipole m = I a area a for loop M = I A area A • Magnetic dipoles one current loop • Evaluate B field along axis passing through loop

B field of large current loop • Consider line integral B. dℓ from loop • Contour C is closed by large semi-circle which contributes zero to line integral I (enclosed by C) z→-∞ a C z→+∞ mo I mo. I/2

Electrostatic potential of dipole sheet • • Now consider line integral E. dℓ from sheet of electric dipoles m = I a I = m/a (density of magnetic moments) Replace I by Np (dipole moment density) and mo by 1/eo Contour C is again closed by large semi-circle which contributes zero to line integral Np/2 eo Electric magnetic -Np/2 eo Field reverses no reversal

Differential form of Ampere’s Law Obtain enclosed current as integral of current density B Apply Stokes’ theorem j dℓ Integration surface is arbitrary S Must be true point wise

Ampere’s Law examples (1) Infinitely long, thin conductor (2) (3) B is azimuthal, constant on circle of radius r (4) Exercise: find radial profile of B inside and outside conductor of radius R B R r B

Solenoid Distributed-coiled conductor Key parameter: n loops/metre B I If finite length, sum individual loops via B-S Law If infinite length, apply Ampere’s Law B constant and axial inside, zero outside Rectangular path, axial length L I L (use label Bvac to distinguish from core-filled solenoids) solenoid is to magnetostatics what capacitor is to electrostatics

Relative permeability Recall how field in vacuum capacitor is reduced when dielectric medium is inserted; always reduction, whether medium is polar or non-polar: is the analogous expression when magnetic medium is inserted in the vacuum solenoid. Complication: the B field can be reduced or increased, depending on the type of magnetic medium

Magnetic vector potential For an electrostatic field We cannot therefore represent B by e. g. the gradient of a scalar since Magnetostatic field, try B is unchanged by