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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 (Wbm-2) or tesla (T) • Magnetostatics in vacuum, then magnetic media based on “magnetic dipole moment” Biot-Savart Law • • I The analogue of Coulomb’s Law is the Biot-Savart Law dB(r) r r-r’ Consider a current loop (I) O r’ dℓ’ • For element dℓ there is an associated element field dB dB perpendicular to both dℓ’ and r-r’ same 1/(4pr2) dependence o is “permeability of free space” defined as 4p x 10-7 Wb A-1 m-1 Integrate to get B-S Law o I d' x(r r' ) dB(r ) 3 4p r r' B(r ) o I d' x(r r' ) 4p r r' 3 B-S Law examples (1) Infinitely long straight conductor dℓ and r, r’ in the page dB 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 dB q p/2 + a sin q = cos a o I B 2p r B r r 2 +z 2 1/2 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 dB is in the page, perpendicular to r - r’, at q to axis. r - r’ I r’ a r z dB q dBz Magnitude of element dB o I d o I d a a dB dB z cosq cosq 2 2 4p r - r ' 4p r - r ' r - r ' a 2 + z 2 1/2 Integrating around loop, only z-components of dB survive The on-axis field is “axial” On-axis field of circular loop dℓ Bon axis dB z o I 4p r - r ' o I 4p r - r ' r - r’ cosq d 2 cosq 2p a 2 I o Ia 2 2 r - r' a Bonaxis 2 limiting cases: z 0 Bon axis 2a z a and Bon axis q r z dBz 3 Introduce axial distance z, where |r-r’|2 = a2 + z2 o I r’ dB o Ia2 2z3 o I a 2 2a +z 2 2 3 2 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 E p 4p or 3 2 cosq rˆ + sinq qˆ om 2 cosq rˆ + sinq qˆ 3 4p r where m p a 2 I a I or m p a 2 I zˆ a area enclosed by current loop B 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) a z→-∞ B.d C a o I 2 a 2dz 2 +z 2 3/2 C a a 2dz 2 2 +z 2 3/2 z→+∞ + 0 (semi circle) o I B.d oI/2 oI 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 o by 1/o Contour C is again closed by large semi-circle which contributes zero to line integral E.d + 0 (semi circle) E.d 0 Electric magnetic Np/2o E.d C -Np/2o Field reverses no reversal Differential form of Ampere’s Law Obtain enclosed current as integral of current density B.d I o encl o j.dS S B Apply Stokes’ theorem B.d B.dS j.dS j dI j.d S o S S dℓ Integration surface is arbitrary B o j Must be true point wise S Ampere’s Law examples (1) Infinitely long, thin conductor B is azimuthal, constant on circle of radius r B.d o Iencl B 2p r o I B B o I 2p r Exercise: find radial profile of B inside and outside conductor of radius R o Ir 2p R 2 I o 2p r Br R B B r R R r 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 B v ac I .d o I encl B v acL o nL I B v ac onI 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: E E vac B rB vac r 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 E.d 0 E - x E x 0 We cannot therefore represent B by e.g. the gradient of a scalar since x B o j (rhs not zero) Magnetostatic field, try also .B 0 always (.E ) o BxA .B . x A 0 x B x x A (see later) B is unchanged by A ' A + x A' x A + x A + 0