Electricity and Magnetism - Physics 121 Lecture 10 - Sources of Magnetic Fields (Currents) Y&F Chapter 28, Sec. 1 - 7 • Magnetic fields are due to currents • The Biot-Savart Law • Calculating field at the centers of current loops • Field due to a long straight wire • Force between two parallel wires carrying currents • Ampere’s Law • Solenoids and toroids • Field on the axis of a current loop (dipole) • Magnetic dipole moment • Summary Copyright R. Janow – Fall 2013 Previously: moving charges and currents feel a force in a magnetic field • Magnets come only as dipole pairs of N and S poles (no monopoles). N S N S N S • Magnetic field exerts a force on moving charges (i.e. on currents). • The force is perpendicular to both the field and the velocity (i.e. it uses the cross product). The F q v B magnetic force can not change a B particle’s speed or KE • A charged particle moving in a uniform magnetic field moves in a circle or a spiral. mv 2 qB R c qB c m • Because currents are moving charges, a wire carrying current in a magnetic field feels a force also using cross product. This force is responsible for the motor effect. FB iLB • For a current loop, the Magnetic dipole moment, torque, and m B potential energy are given by: N i A nˆ Copyright R. Janow – Fall 2013 Um B Magnetic fields are due to currents Oersted - 1820: A magnetic compass is deflected by current Magnetic fields are due to currents (free charges & in wires) In fact, currents are the only way to create magnetic fields. The magnitude of the field created is proportional to i s (current-length) Copyright R. Janow – Fall 2013 The Biot-Savart Law (1820) • Same basic role as Coulomb’s Law: magnetic field due to a source dB • Source strength measured by “current-length” i ds • Falls off as inverse-square of distance ids x r • New constant 0 measures “permeability” • Direction of B field depends on a cross-product (Right Hand Rule) Differential addition to field at P due to distant source ids Unit vector P’ idsrˆ along r - 0 dB dB from source to P 4 r2 (out of page) 10-7 exactly “vacuum permeability” -7 0 = 4x10 T.m/A. Find total field B by integrating over the whole current region (need lots of symmetry) For a straight wire the magnetic field lines are circles wrapped around it. Another Right Hand Rule shows the direction: ids sin() dB 0 4 r2 Copyright R. Janow – Fall 2013 Direction of Magnetic Field 10 – 1: Which sketch below shows the correct direction of the magnetic field, B, near the point P? B into B into B into page page B B page i i i i i P P P P P A B C D E Use RH rule for current segments: thumb along ids - curled fingers show B Copyright R. Janow – Fall 2013 Example: Magnetic field at the center of a current arc • Circular arc carrying current, constant radius R • Find B at center, point C • f is included arc angle, not the cross product angle • Angle for the cross product is always 900 • dB at center C is up out of the paper • ds = Rdf’ i 0 ds 0 d' dB i i B 0 idsrˆ 4 R2 4 R dB R ids R • integrate4 on arcr2 angle f’ from 0 to f f Right hand rule 0i 0i for wire segments B df' f f in radians 4R 4R 0 • For a circular loop of current - f = 2 radians: 0i B B B (loop) 2R Thumb points along the current. Curled fingers Another Right Hand Rule (for loops): show direction of B Curl fingers along current, thumb shows direction of B at center Copyright R. Janow – Fall 2013 ?? What would formula be for f = 45o, 180o, 4 radians ?? Examples: FIND B FOR A POINT LINED UP WITH A SHORT STRAIGHT WIRE i ds P i dsrˆ 0 dB 0 i ds sin() 0 rˆ 4r2 Find B AT CENTER OF A HALF LOOP, RADIUS = r i i B 0 0 4r 4r into page Find B AT CENTER OF TWO HALF LOOPS 0i 0i OPPOSITE B 2 x CURRENTS 4r 2r same as closed loop 0i 0i PARALLEL B - 0 CURRENTS 4r 4r into out of Copyright R.page Janow – Fallpage 2013 Magnetic Field from Loops 10 – 2: The three loops below have the same current. The smaller radius is half of the large one. Rank the loops by the magnitude of magnetic field at the center, greatest first. A. I, II, III. B. III, I, II. C. II, I, III. D. III, II, I. E. II, III, I. I. II. III. 0i B f f in radians Hint: consider radius, direction, arc angle 4R Copyright R. Janow – Fall 2013 Magnetic field due to current in a thin, straight wire ids r • Current i flows to the right along x – axis dB 0 • Wire subtends angles 1 and 2 3 • Find B at point P, a distance a from wire. 4 r • dB is out of page at P for ds anywhere along wire a Evaluate dB along ds using Biot Savart Law • Magnitude of i.ds X r = i.r.dx.cos(). i dx cos() dB 0 kˆ 4 r2 • x negative as shown, positive, 1 positive, 2 negative d r a /cos() x atan() [tan()] sec2 () 1/ cos2 () d dx a d/cos2 () i | dB | 0 cos()d 4a Integrate on from 1 to 2: 2 0i 2 0i B dB cos()d [sin(1) sin(2 )] 1 4a 1 4a General result – applications follow Copyright R. Janow – Fall 2013 Magnetic field due to current in thin, straight wires i B 0 [sin( ) sin( )] 4a 1 2 Example: Infinitely long, thin wire: Set 1 /2, 2 /2 [ direction was CW in sketch ] RIGHT HAND RULE FOR A WIRE i B 0 2a a is distance perpendicular to FIELD LINES ARE CIRCLES wire through P THEY DO NOT BEGIN OR END Example: Field at P due to Semi-Infinite wires: Set /2, 0 1 2 Into slide at point P Half the magnitude for a Zero i fully infinite wire contribution | B | 0 4a Copyright R. Janow – Fall 2013 Magnetic Field lines near a straight wire carrying current i out of slide When two parallel wires are carrying current, the magnetic i field from one causes a force on the other. Fa,b ibLb Ba 0ia Ba . The force is attractive when the 2R currents are parallel. The force is repulsive when the currents are anti-parallel. Copyright R. Janow – Fall 2013 Magnitude of the force between two long parallel wires L • Third Law says: F12 = - F21 i1 • Use result for B due to infinitely long wire i x x x x x B 0 1 Due to 1 at wire 2 d x x x x x 1 2d Into page via RH rule x x x x x i2 • Evaluate F12 = force on 2 due to field of 1 x x x x x F12 i2L2 B1 x x x x x i2L is normal to B Force is toward wire1 | F1,2 | i2L B1 End View i i F 0 1 2 L F = - F 1,2 2 d 21 12 i1 • Attractive force for parallel currents i 2 • Repulsive force for opposed currents Example: Two parallel wires are 1 cm apart |i1| = |12| = 100 A. 2x10-7 x 100 x 100 F/L force per unit length 0.2 N / m .01 F 0.2 N for L 1 m Copyright R. Janow – Fall 2013 Forces on parallel wires carrying currents 10 – 3: Which of the four situations below results in the greatest force to the right on the central conductor? The currents in all the wires have the same magnitude. i B 0 F i LB greatest F ? 2R tot tot 1 2 3 4 A. B. C. D. Hints: Which pairings with center wire are attractive and repulsive? or What is the field midway between wires with parallel currents? What is the net field directions and relaative magnitudes atCopyright center wireR. Janow – Fall 2013 Ampere’s Law • Derivable from Biot-Savart Law • Sometimes a way to find B, given the current that creates it • But B is inside an integral usable only for high symmetry (like Gauss’ Law) • An “Amperian loop” is a closed path of any shape ienc= net current • Add up (integrate) components Bds 0ienc passing through the of B along the loop path. loop To find B, you have to be able to do the integral, then solve for B Picture for applications: • Only the tangential component of B along ds contributes to the dot product • Current outside the loop (i3) creates field but doesn’t contribute to the path integral • Another version of RH rule: - curl fingers along Amperian loop - thumb shows + direction for net current Copyright R. Janow – Fall 2013 Example: Find magnetic field outside a long, straight, possibly fat, cylindrical wire carrying current i We used the Biot-Savart Law to show that B 0 for a thin wire 2r Now use Ampere’s Law to show it again more simply and for a fat wire. B ds 0ienc Amperian loop outside R can have any shape Choose a circular loop (of radius r>R) because field lines are circular about a wire. B and ds are then parallel, and B is constant R everywhere on the Amperian path B ds Bx2r i 0 enc The integration was simple.