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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 for wire element and moving charge 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 2015 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). • Magnetic field exerts a force on moving charges (i.e. on currents). • The Lorentz force is perpendicular to both the field and the velocity Magnetic force can not change a particle’s speed or KE • A charged particle moving in a uniform magnetic field moves in a circle or a spiral. 2 qB mv c R c m qB • A wire carrying current in a magnetic field feels a force also using cross product. This force is the motor effect. • • Current loops behave as Magnetic dipoles, with dipole moment, torque, and potential energy given by: N S N S NS FB q v B FB iL B N i A n̂ dF i ds B m B Um B Copyright R. Janow – Fall 2015 Magnetic fields are created by currents Oersted - 1820: A magnetic compass is deflected by current Magnetic fields are due to currents (free charges & in wires) Why do the un-magnetized filings line up with the field? In fact, currents are the only way to create magnetic fields. The magnitude of the field created is proportional to i s or qv (current - length) Copyright R. Janow – Fall 2015 Biot-Savart Law (1820): Field due to a current element • Same basic role as Coulomb’s Law: magnetic field due to a source • Source strength measured by “current-length” i ds • Falls off as inverse-square of distance • 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 current element ids 0 id s r̂ dB 4 r 2 10-7 exactly Unit vector along r from source to P id s dB x r P’ dB (out of page) “vacuum permeability” 0 = 4x10-7 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 2015 Magnetic field due to a moving point charge • A variant of Biot-Savart Law • Source is qv having dimensions of current-length • Magitude of B falls off as inverse-square of distance • Direction of B field depends on a cross-product (Right Hand Rule) qv dB r x Field at P due to distant source qv qv r̂ 0 B 4 r 2 Unit vector along r from source to P 10-7 exactly r̂ The magnetic field lines are circles wrapped around the current length qv. r̂ Velocity into page B 0 qv sin() 4 r2 Copyright R. Janow – Fall 2015 Direction of Magnetic Field 10 – 1: Which sketch below shows the correct direction of the magnetic field, B, near the point P? B into page i P A B into page B P i i B P C Use RH rule for current segments: thumb along ids - curled fingers show B i B B into page P P D i E 0 id s r̂ dB 4 r 2 Copyright R. Janow – Fall 2015 Example: Magnetic field at the center of a current arc • • • • • • Circular current arc, radius R Find B at center, point C 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 = Rd’ i B 0 ds 0 d' dB i 2 i 4 R 4 R • integrate on arc angle ’ from 0 to 0i 0i B d' 4R 0 4R in radians ids R 0 id s r̂ dB 4 r 2 R Right hand rule for wire segments • For a circular loop of current - = 2 radians: B B 0i B (loop) 2R Thumb points along the current. Curled fingers show direction of B Another Right Hand Rule (for current loops only): Curl fingers along current, thumb shows direction of B at center ?? What would formula be for = 45o, 180o, Copyright R. Janow – Fall 2015 4 radians ?? Examples: FIND B FOR A POINT LINED UP WITH A SHORT STRAIGHT WIRE 0 i ds P dB i ds sin() i ds r̂ 0 2 r̂ 4r 0 Find B AT CENTER OF A HALF LOOP, RADIUS = r B 0i 0i 4r 4r into page Find B AT CENTER OF TWO HALF LOOPS OPPOSITE CURRENTS 0i 0i B 2x 4r 2r same as closed loop PARALLEL CURRENTS B 0i 0i 0 4r 4r into page out of page Copyright R. Janow – Fall 2015 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. B. C. D. E. I, II, III. III, I, II. II, I, III. III, II, I. II, III, I. 0i B 4R I. in radians II. III. Hint: consider radius, direction, arc angle Copyright R. Janow – Fall 2015 Magnetic field due to current in a thin, straight wire • Current i flows to the right along x – axis • Wire subtends angles 1 and 2 at P • Find B at point P, a distance a from wire. ids r̂ 0 dB 4 r 2 • dB points out of page at P for ds anywhere along wire a Evaluate dB due to ids using Biot Savart Law ids i î dx i dx cos() dB 0 k̂ 4 r2 • Angles , 1, 2 measured CW about –y axis • x negative as shown, positive, 1 positive, 2 negative d x a tan() r a / cos() [tan()] sec 2 () 1/ cos2 () d dx a d(tan()) / d a d / cos2 () i | dB | 0 cos()d 4 a Integrate on from 1 to 2: i B | dB | 0 1 4a 2 2 1 Symmetric Case: cos()d 0i [sin(1) sin(2 )] 4a 2 1 B 0i sin(1) 2a Copyright R. Janow – Fall 2015 Magnetic field due to current in thin, straight wires B 0i [sin(1 ) sin(2 )] 4 a Special Case: Infinitely long, thin wire: Set 1 / 2, 2 / 2 [ direction was CW in sketch ] RIGHT HAND RULE FOR A WIRE 0i B 2a a is distance perpendicular to wire through P FIELD LINES ARE CIRCLES THEY DO NOT BEGIN OR END Example: Field at P due to Semi-Infinite wires: A Set 1 / 2, 2 0 (piece B) Zero contribution B 0i |B| 4 a Field is into slide at point P Half the magnitude for a fully infinite wire Copyright R. Janow – Fall 2015 Magnetic Field lines near a straight wire carrying current i out of slide i When two parallel wires are carrying current, the magnetic field from one causes a force on the other. 0ia Ba 2R Fon b ibLb Ba . The force is attractive when the currents are parallel. . The force is repulsive when the Copyright R. Janow – Fall 2015 currents are anti-parallel. Magnitude of the force between two long parallel wires L i1 d • Third Law says: F12 = - F21 • Use result for B due to infinitely long wire x x x x x x x x x x i2x x x x x x x x x x x x x x x End View 0i1 B1 2d • Evaluate F12 = force on 2 due to field of 1 Fon 2 i2L2 B1 | Fon 2 | i2L B1 F1,2 i1 i2 Example: Due to 1 at wire 2 Into page via RH rule 0 i1i2 L 2 d i2L is normal to B Force is toward wire1 F21= - F12 • Attractive force for parallel currents • Repulsive force for opposed currents 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 2015 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. 0i B 2R A. 1 Ftot i L B tot 2 greatest F ? 3 4 B. C. D. Answer B Hints: Which pairings with center wire are attractive, repulsive, or zero? What is the field midway between wires with parallel currents? What is the net field direction and relative magnitude at center wire Copyright R. Janow – Fall 2015 Ampere’s Law • Derivable from Biot-Savart Law – A theorem • 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) • Integrate over an “Amperian loop” - a closed path. • Adds up components of B along the loop path. • Any path enclosing a set of currents has same result. B ds 0ienc ienc= net current passing through the loop 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 2015 Ampere’s Law Example: Find magnetic field outside a long, straight, possibly fat, cylindrical wire carrying current The Biot-Savart Law showed that for a thin infinitely long wire: B 0i 2r Now Ampere’s Law shows it again more simply and for a long fat wire. End view of wire with radius R B ds 0ienc Amperian loop outside R can have any shape Choose a circular loop (of radius r>R) because field lines are circular around a wire. B and ds are parallel and B is constant everywhere on the Amperian path R B ds Bx2r 0ienc The integration was simple. ienc is the total current. Solve for B to get our earlier expression: 0i B 2r outside wire R has no effect on the result. (reminiscent of Shell Theorem) Copyright R. Janow – Fall 2015 Magnetic field inside a long straight wire carrying current, via Ampere’s Law B ds 0ienc Assume current density J = i/A is uniform across the wire cross-section and is cylindrically symmetric. Field lines are again concentric circles B is axially symmetric again Again, choose a circular Amperian loop path around the axis, of radius r < R. The enclosed current is less than the total current i, because some is outside the Amperian loop. The amount enclosed is ienc i r 2 R 2 Apply Ampere’s Law: r2 B ds B2r 0ienc 0i R 2 i r B 0 2R R r R inside wire B ~r ~1/r R r Outside (r > R), the wire looks like an infinitely thin wire (previous expression) Copyright R. Janow Fall Inside: B(r) grows linearly up –to R2015 Counting the current enclosed by an Amperian Loop B ds 10 – 4: Rank the Amperian paths shown by the value of along each path, taking direction into account and putting the most positive ahead of less positive values. All of the wires are carrying the same current.. A. I, II, III, IV, V. B. II, III, IV, I, V. C. III, V, IV, II, I. D. IV, V, III, I, II. E. I, II, III, V, IV. B ds 0ienc I. II. III. IV. V. Copyright R. Janow – Fall 2015 Another Ampere’s Law example Why use COAXIAL CABLE for CATV and other applications? Find B inside and outside the cable Cross section: B ds 0ienc shield wire current i into sketch Amperian loop 1 Amperian loop 2 center wire current i out of sketch Field Inside – use Amperian loop 1: B ds 0i Bx2r 0i B 2r Field Outside – use Amperian loop 2: B ds 0ienc 0 B0 Outer shield current does not affect field inside Reminiscent of Gauss’s Law Zero field outside due to opposed currents + radial symmetry Copyright R. Janow – Fall 2015 Losses and interference suppressed Solenoids strengthen fields by using many loops “Long solenoid” d << L L d strong uniform field in center n # coils / unit length N/L Approximation: field is constant inside and zero outside (just like capacitor) fields from adjacent coils cancel FIND FIELD INSIDE IDEAL SOLENOID USING AMPERIAN LOOP abcda B ds 0ienc B ds Binsideh 0ienc 0inh only section that has nonzero contribution • Outside B = 0, no contribution from path c-d • B is perpendicular to ds on paths a-d and b-c • Inside B is uniform and parallel to ds on path a-b B 0in inside ideal solenoid Copyright R. Janow – Fall 2015 Toroid: A long solenoid bent into a circle i outside flows up Find the magnitude of B field inside LINES OF CONSTANT B ARE CIRCLES Draw an Amperian loop parallel to the field, with radius r (inside the toroid) The toroid has a total of N turns The Amperian loop encloses current Ni. B is constant on the Amperian path. B ds B 2r 0ienc 0iN 0iN B 2 r inside toroid • N times the result for a long thin wire • Depends on r • Also same result as for long solenoid n N 2 r (turns/uni t length) B 0in Find B field outside (ienc=0) AMPERIAN LOOP IS A CIRCLE ALONG B Answer B0 outside Copyright R. Janow – Fall 2015 Find B at point P on z-axis of a current loop (dipole) • Use the Biot-Savart Law directly i ds r̂ dB 0 4 r 2 dB r R 2 z2 cos R r cancels opposite side by symmetry (normal to z - axis) Need only component along z 0 i ds cos() dBz dB|| dB cos() 4 R 2 z2 iR ds R d dBz 0 2 ds 2 3 / 2 4 (R z ) • • Integrate around the current loop on – the angle at the center of the loop. The field is perpendicular to r but by symmetry the component of B normal to z-axis cancels around the loop - only the part parallel to the z-axis survives. Bz dBz B(z) 0 0 iR iR 2 ds 4 (R 2 z 2 ) 3 / 2 4 (R 2 z 2 )3 / 2 0i R 2 2(R 2 z 2 )3 / 2 as before i B(z 0) 0 2R 2 0 d i is into page recall definition of Dipole moment 2 NiA Ni–RFall Copyright R. Janow 2015 B field on the axis of a dipole (current loop), continued Far field: suppose z >> R B(z) 0iR 2 2(R z ) 2 2 3/ 2 Same 1/z3 dependence as for electrostatic dipole 0iR 2 2z 3 is normal to loop (RH Rule). Dipole moment vector ˆ NiA N number of turns 1 A area of loop R 2 R2i | | above For any current loop, along z axis with |z| >> R 0 B(z) 2 z 3 For charge dipole E(z) p 1 2 0 z 3 Current loops are the elementary sources of magnetic field: • Create dipole fields with source strength • Dipole feels torque due to another in external B field B Dipole-dipole interaction: 1 r3 2 Torque 1 2 depends on Copyright R. Janow – Fall 2015 Try this at home 10-5: The three loops below have the same current. Rank them in terms of the magnitude of magnetic field at the point shown, greatest first. A. I, II, III. B. III, I, II. C. II, I, III. D. III, II, I. E. II, III, I. I. B 0i 4R II. III. Hint: consider radius, direction, arc angle in radians Answer: B Copyright R. Janow – Fall 2015