Magnetic fields are due to currents

1. 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 a moving charge and for a wire element • 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

2. 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 magnetic force is perpendicular to both the field and the velocity Magnetic force can not change a particle’s speed or KE N S N S N S • A charged particle moving in a uniform magnetic field moves in a circle or a spiral. • A wire carrying current in a magnetic field feels a force, also summarized by a cross product. • This force is the motor effect. • Current loops behave as magnetic dipoles, with dipole moment, torque, and potential energy given by: Previously: moving charges and currents feel a force in a magnetic field

3. Electromagnetic crane Why do the un-magnetized filings line up with the field? Now: Currents Create Magnetic Field Oersted - 1820: A magnetic compass is deflected by current  Magnetic fields are due to currents (free charges & in conductors) In fact, currents are the only way to create magnetic fields. The magnitude of the field created is proportional to

4. x Field at P due to distant source qv Snapshot. Time dependent Unit vector along r - from source to P 10-7 exactly The magnetic field lines are circles wrapped around the current length qv. Velocity into page Magnetic field due to a moving point charge • Variant of Biot-Savart Law • “Current length vector” qv is a point source for field • Magnitude of B falls off as inverse-square of distance • Direction of B is via cross-product (Right Hand Rule) • Field at a point changes as particle moves past

5. x Differential addition to field at P due to distant source current element ids Unit vector along r - from source to P P’ “vacuum permeability” m0 = 4px10-7 T.m/A. 10-7 exactly Integrate dB over an extended current region to find total field at point P. Usually need lots of symmetry to do the integration. For a straight wire the magnetic field lines are circles wrapped around it. Another Right Hand Rule shows the direction: Biot-Savart Law (1820): Field due to current in a wire • Same basic role as Coulomb’s Law: magnetic field due to a source • “Current-length vector” i ds is a point source for B field • Field falls off as inverse-square of distance • New constant m0 measures “vacuum permeability” • Direction of B field depends on a cross-product (right hand rule)

6. B into page B into page B into page B B i i i i P P P P P i A B C D E Magnetic Field Direction near a Straight Wire 10 – 1: Which sketch below shows the correct direction of the magnetic field, B, created near point P by current i? Use RH rule for current segments: thumb along ids - curled fingers show B

7. i • Integrate on arc angle f’ from 0 to f Right hand rule for wire segments • Apply to circular current loop - f = 2p radians: 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 f = 45o, 180o, 4p radians ?? Example: Magnetic field at the center of a current arc • Circular current arc, radius R • Find B at center, point C • f is included arc angle, not the cross product angle • Cross product angle q is always 900 here • dB at center C is toward us • ds = Rdf

8. Examples: P Find B AT CENTER OF A HALF LOOP, RADIUS = r into page Find B AT CENTER OF TWO HALF LOOPS OPPOSITE EQUAL CURRENTS same as closed loop PARALLEL EQUAL CURRENTS into out of page page FIND B FOR A POINT LINED UP WITH A SHORT STRAIGHT WIRE

9. I. II. III. Magnetic field at the center of current 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. • I, II, III. • III, I, II. • II, I, III. • III, II, I. • II, III, I. Hint: consider radius, direction, arc angle

10. Magnetic field near a long, thin straight wirecarrying current a is distance perpendicular to wire through P L = length of wire >> a Field of infinitely long wire RIGHT HAND RULE FOR A WIRE Magnetic field lines are concentric circles around wire FIELD LINES ARE CIRCLES THEY DO NOT BEGIN OR END Above formula derived by Ampere’s Law (below) or direct integration along finite-length wire Crops up as an approximation in problems (as did point charge for electrostatics) Field intensity falls off as 1/a not 1/a2

11. i out of slide i RIGHT HAND RULE FOR A WIRE i FIELD LINES ARE CIRCLES THEY DO NOT BEGIN OR END Magnetic field lines near a straight wire carrying current

12. Example: Force between two parallel wires carrying current The magnetic field from one wire causes a force on the other. . The force is attractive when the currents are parallel.. The force is repulsive when the currents are anti-parallel.

13. L ia x x x x x x x x x x x x x x x x x x x x d ib • Evaluate Fab = force on b due to field of a ibL is normal to Ba Force is toward wire a x x x x x Fab= - Fba ia • Attractive force / unit length for parallel • currents • Repulsive force for opposed currents ib Example: Two parallel wires are 1 cm apart |i1| = |i2| = 100 A. Magnitude of the force between two long parallel wires • Third Law says: Fab = - Fba • Use formula for field of infinitely long wire Field of wire a at wire b Into page via RH rule End View

14. 1 2 3 4 A. B. C. D. At home: forces on parallel wires carrying currents 10 – 3: Which situation below results in zero force on the center wire? The currents in all the wires have the same magnitude. Zero force if Bnet = 0 Answer A Hints: What is the field midway between wires for parallel & opposite currents? What is the net field direction and relative magnitude at center wire

15. Line Integral over an “Amperian • loop” - a closed path. • Adds up components of B • along the loop path. • All paths that enclose a certain set of currents have same result. 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 contributions to the path integral • cancel around closed path. • Another version of RH rule: • - curl fingers along Amperian loop • - thumb shows + direction for net current Ampere’s Law • Derivable from Biot-Savart Law – A theorem (not proven here) • 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)

16. 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 The path integration simplifies. ienc is the total current i. Solve for B, get same formula as for the thin wire: R has no effect on the result as long as R < r. (reminiscent of Shell Theorem) Example: Find magnetic field outside a long, straight, wire (or cylinder) carrying current for a thin infinitely long wire: The Biot-Savart Law showed that Now Ampere’s Law proves it again more simply for a long fat wire. End view of wire with radius R

17. B ~r Apply Ampere’s Law: ~1/r r R Outside (r > R), the wire looks like an infinitely thin wire (previous expression) Inside: B(r) grows linearly up to R Magnetic field inside a long straight wire carrying current, via Ampere’s Law Assume cylindrical symmetry Assume current density J = i/A is uniform across the wire cross-section. Field lines inside wire are again concentric circles B is cylindrically symmetric again Again, choose a circular Amperian loop path around the axis, of radius r < R. The enclosedcurrent is less than the total current i, because some current is outside the Amperian loop. The amount enclosed is:

18. 10 – 4: Rank the Amperian paths shown by the value of along each path in descending order, starting with the most positive, taking current direction into account. The current is the same in all of the wires. I. II. III. IV. V. At home: count the current enclosed by an Amperian Loop • I, II, III, IV, V. • II, III, IV, I, V. • III, V, IV, II, I. • IV, V, III, I, II. • I, II, III, V, IV. Answer: E

19. 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: Outer shield current does not affect field inside Reminiscent of Gauss’s Law Field Outside – use Amperian loop 2: Zero field outside due to opposed currents + radial symmetry Losses and interference suppressed Another Ampere’s Law Example Why use COAXIAL CABLE for CATV and other applications? Find B inside and outside the cable Cross section:

20. L “Long solenoid”  d << L d strong uniform field in center Approximation: field is constant inside and zero outside (just like capacitor) fields from adjacent coils cancel outside FIND FIELD INSIDE IDEAL SOLENOID USING AMPERIAN LOOP abcda only section that has non-zero contribution inside ideal solenoid • Outside B = 0, no contribution from path c-d • B is perpendicular to ds on paths d-a and b-c • Inside B is uniform and parallel to ds on path a-b Solenoids strengthen fields by having many current loops

21. Inversely proportional to r • Same result as for long solenoid • N times the result for a long thin wire Find B field outside (ienc=0) Answer A Toroid is a long solenoid bent into a circle Find the magnitude of B inside • Choose a different Amperian loop, parallel to the field, circular, 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. i outside flows up LINES OF CONSTANT B ARE CIRCLES AMPERIAN LOOP IS A CIRCLE ALONG B

22. f is the azimuth angle at the center of the loop. • Integrate on azimuth f around the current 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. i is into page as before recall definition of Dipole moment Example: B field at point P on z-axis of a current loop (dipole) • Use the Biot-Savart Law to directly integrate

23. Far field: suppose z >> R Same 1/z3 dependence as for field of electrostatic dipole Dipole moment vector is normal to loop (RH Rule). For any current loop, on z axis with |z| >> R For charge dipole • Current loops are the elementary sources of magnetic field: • Dipole 1 creates field proportional to source strength • Dipole 2 feels torque proportional to in external B field Dipole-dipole interaction: Torque depends on r 3 B field on the axis of a dipole (current loop), continued

24. Quick summary of magnetism so far Current-lengths in a magnetic field feel forces and torques Force on charge and wire carrying current torque and potential energy of a dipole Current loops are elementary dipoles Biot-Savart Ampere q B on the symmetry axis of a current loop (far field): B at center of curvature of a circular arc of current: Current-lengths (changing electric fields) produce magnetic fields B outside a long straight wire carrying a current i: B at center of circular loop carrying a current i : Long Solenoid field: B inside a torus carrying B outside = 0 a current i :

25. SUPPLEMENTARY MATERIAL

26. EXTRA: Magnetic field due to current in a thin finite length wire, • Current i flows to the right along x – axis • Wire subtends angles q1 and q2 at P • Find B at point P, a distance a from wire. • dB points out of page at P for ds anywhere along wire Evaluate dB due to current ids using Biot Savart Law • Angles q, q1, q2 measured CW from –y axis • x shown is negative, q positive, q1 positive, q2 negative Integrate on q from q1 to q2: Symmetric Case: a

27. Magnetic field near current-carrying thin, straight wire Example: Field at P due to Semi-Infinite wires A & B: A Field is into slide at point P Half the magnitude found for a fully infinite wire B Zero Contribution From A FIELD LINES ARE CIRCLES THEY DO NOT BEGIN OR END Special Case: Infinitely long, thin wire. Use result above: RIGHT HAND RULE FOR A WIRE a is distance perpendicular to wire through P

28. Denominator has same sign for s positive and negative • Substitute: • From integral • tables: RIGHT HAND RULE FOR A WIRE R is distance perpendicular to wire through P FIELD LINES ARE CIRCLES THEY DO NOT BEGIN OR END Magnetic field near infinitely long, straight wire • Wire is infinitely long, current i flows up • Find B at point P.dB is into page at P for dsanywhere along wire • Magnitude of B is the same everywhere on any circle of radius R • Magnitude of ids X r = irds sin(q). • Integrate dB using Biot Savart Law from s = –infinity to +infinity

29. 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 center of curvature, greatest first. • I, II, III. • III, I, II. • II, I, III. • III, II, I. • II, III, I. I. II. III. Hint: consider radius, direction, arc angle Answer: B