Sources of the Magnetic Field March 20 2008 Note These slides will be updated for the actual presentation Remember the wire Try to remember 1 r dq 1 rdq dE 2 4 0 r r 4 0 r 3 r UNIT VECTOR r The Coulomb s Law of Magnetism A Vector Equation duck For the Magnetic Field current elements create the field This is the Law of Biot Savart In a similar fashion to E field 0 ids runit 0 ids r B 3 2 4 r 4 r permeability 0 4 10 7 Tm A 1 26 10 7 Tm Magnetic Field of a Straight Wire We intimated via magnets that the Magnetic field associated with a straight wire seemed to vary with 1 d We can now PROVE this From the Past Using Magnets Right hand rule Grasp the element in your right hand with your extended thumb pointing in the direction of the current Your fingers will then naturally curl around in the direction of the magnetic field lines due to that element Let s Calculate the FIELD Note For ALL current elements in the wire ds X r is into the page The Details 0 ids sin dB 4 r2 Negative portion of the wire contributes an equal amount so we integrate from 0 to and DOUBLE it 0i sin ds B 2 2 r 0 Moving right along 2 r s R 2 sin sin R s2 R2 So 0i 0i Rds B 3 2 2 0 s 2 R 2 2 R 1 d A bit more complicated A finite wire P1 NOTE sin sin ds r ds r sin r ds R sin r 0i ds sin dB 2 4 r 2 r s R 2 1 2 More P1 L 2 0i ds B 3 2 2 2 4 L 2 s R and 0i L B 2 R L2 4 R 2 when L 0i B 2 R P2 0 0iR ds B 2 2 3 2 4 L s R or 0i L B 4 R s 2 R 2 APPLICATION point P Find the magnetic field B at Center of a Circular Arc of a Wire carrying current More arc ds ds Rd 0 ids 0 iRd dB 2 4 R 4 R 2 0 iRd 0i B dB d 2 4 R 4 R 0 0 0 i B at point C 4 R The overall field from a circular current loop Sorta looks like a magnet Iron Howya Do Dat ds r 0 No Field at C Force Between Two Current Carrying Straight Parallel Conductors Wire a creates a field at wire b Current in wire b sees a force because it is moving in the magnetic field of a The Calculation The FIELD at wire b due to wire a is what we just calculated 0ia Bat b 2 d Fon b ib L B Since L and B are at right angles 0 Lia ib F 2 d Definition of the Ampere The force acting between currents in parallel wires is the basis for the definition of the ampere which is one of the seven SI base units The definition adopted in 1946 is this The ampere is that constant current which if maintained in two straight parallel conductors of infinite length of negligible circular cross section and placed 1 m apart in vacuum would produce on each of these conductors a force of magnitude 2 x 10 7 newton per meter of length Ampere s Law The return of Gauss Remember GAUSS S LAW qenclosed E d A 0 Surface Integral Gauss s Law Made calculations easier than integration over a charge distribution Applied to situations of HIGH SYMMETRY Gaussian SURFACE had to be defined which was consistent with the geometry AMPERE S Law is the Gauss Law of Magnetism Sorry The next few slides have been lifted from Seb Oliver on the internet Whoever he is Biot Savart The Coulombs Law of Magnetism 0 ids r dB 2 4 r Invisible Summary Biot Savart Law ids r dB 0 2 4 r 0 I produced by wires B 2R Centre of a wire loop radius R 0 NI B Centre of a tight Wire Coil with N turns 2R Distance a from long straight wire Field Force between two wires Definition of Ampere 0 I B 2 a F 0 I1 I 2 l 2 a Magnetic Field from a long wire Using Biot Savart Law r I B ds Take a short vector on a circle ds B ds B ds cos 0 cos 1 Thus the dot product of B the short vector ds is 0 I B 2 r B ds B ds 0 I B ds ds 2 r Sum B ds around a circular path r 0 I B ds ds 2 r I B Sum this around the whole ring ds Circumference of circle B ds ds 2 r 0 I ds 2 r 0 I ds 2 r 0 I B ds 2 r 0 2 r Consider a different path B ds 0 Field goes as 1 r Path goes as r Integral independent of r i SO AMPERE S LAW by SUPERPOSITION We will do a LINE INTEGRATION Around a closed path or LOOP Ampere s Law B d s i 0 enclosed USE THE RIGHT HAND RULE IN THESE CALCULATIONS The Right Hand Rule AGAIN Another Right Hand Rule COMPARE B d s i 0 enclosed Line Integral qenclosed E d A 0 Surface Integral Simple Example Field Around a Long Straight Wire B ds i 0 enclosed B 2 r 0i 0i B 2 r Field INSIDE a Wire Carrying UNIFORM Current The Calculation B ds B ds 2 rB i 0 enclosed ienclosed r 2 i 2 R and 0i B r 2 2 R B 0i 2 R R r Procedure Apply Ampere s law only to highly symmetrical situations Superposition works Two wires can be treated separately and the results added VECTORIALLY The individual parts of the calculation can be handled usually without the use of vector calculations because of the symmetry THIS IS SORT OF LIKE GAUSS s LAW WITH AN ATTITUDE The figure below shows a cross section of an infinite conducting sheet carrying a current per unit x length of l the current emerges perpendicularly out of the page a Use the Biot Savart law and symmetry to show that for all points P above the sheet and all points P below it the magnetic field B is parallel to the sheet and directed as shown b Use Ampere s law to find B at all points P and P FIRST PART Vertical Components Cancel Apply Ampere to Circuit L B Infinite Extent B current per unit length Current inside the loop is therefore i L The Math 0 s d B B Infinite Extent B B ds i 0 enclosed BL BL 0 L 0 B 2 A Physical Solenoid Inside the Solenoid For an INFINITE long solenoid the previous problem and SUPERPOSITION suggests that the field OUTSIDE this solenoid is ZERO More on Long …
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