Sources of Magnetic Fields (Currents) Y&F Chapter 28, Sec

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

• 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  iLB  • 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’  idsrˆ along r -  0 dB dB  from source to P 4 r2 (out of page)

10-7 exactly “vacuum permeability” -7 0 = 4x10 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

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   idsrˆ dB  0 • Find B at center, point C 2 • f is included arc angle, not the cross product angle 4 r • 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 4 2 4    R R R ids  R • integrate on arc angle f’ from 0 to f f Right hand rule 0i 0i for wire segments B  df'  f f in radians 4R  4R 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 dsrˆ  0 dB  0 i ds sin()  0 rˆ 4r2 Find B AT CENTER OF A HALF LOOP, RADIUS = r

 i  i B  0   0 4r 4r into page Find B AT CENTER OF TWO HALF LOOPS

0i 0i OPPOSITE B  2 x   CURRENTS 4r 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 4R

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 4a

Integrate on  from 1 to 2:

  2 0i 2 0i B  dB   cos()d  [sin(1)  sin(2 )]   1 4a 1 4a

General result – applications follow Copyright R. Janow – Fall 2013 Magnetic field due to current in thin, straight wires  i B  0 [sin( )  sin( )] 4a 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 2a

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 4a 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 2R 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 2d 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 LB greatest F ? 2R tot tot

1 2 3 4 A.

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 Bds  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 2r 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  Bx2r   i  0 enc

The integration was simple. ienc is the total current. Solve for B to get our earlier expression:

0i B  outside wire R has no effect on the result. 2r Copyright R. Janow – Fall 2013 Magnetic field inside a long straight   B ds   i wire carrying current, via Ampere’s Law  0 enc

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 draw a circular Amperian loop 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 r2 ienc  i R2 B ~r Apply Ampere’s Law: ~1/r   r2 B ds  B2r  0ienc  0i  R2 R r Outside (r>R), the wire looks like an   i  r  B   0   r  R inside wire infinitely thin wire (previous expression)  2R  R  Inside: B growsCopyright linearly R. upJanow to –R Fall 2013 Counting the current enclosed by an Amperian Loop   10 – 4: Rank the Amperian paths shown by the value of  B ds 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. I. D. IV, V, III, I, II. E. I, II, III, V, IV. II.

  III. B ds  0ienc IV. V.

Copyright R. Janow – Fall 2013 Another Ampere’s Law example Why use COAXIAL CABLE for CATV and other applications? Find B inside and outside the cable Cross section:

Amperian shield wire loop 1 current i into sketch Amperian loop 2 center wire current i out of sketch

Inside – use Amperian loop 1:   Outer shield does not 0i B ds  0i  Bx 2r B  affect field inside 2r Reminiscent of Gauss’s Law Outside – use Amperian loop 2: Zero field outside due to opposed   B ds   i  0 currents + radial symmetry  0 enc Losses and interferenceCopyright R. Janow suppressed – Fall 2013 Solenoids strengthen fields by using many loops

L cancellation

n  # coils / unit length  N/L strong uniform Approximation: field is constant inside and field in center zero outside (just like capacitor) “Long solenoid”  d << L

FIND FIELD INSIDE IDEAL SOLENOID USING AMPERIAN LOOP abcda   B ds  0ienc   B ds  Binsideh  0ienc  0inh B   in only section that has non- 0 zero contribution inside ideal solenoid • 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 Copyright R. Janow – Fall 2013 Toroid: A long solenoid bent into a circle Find the magnitude of B field inside i outside . Draw an Amperian loop parallel to the field, with flows up radius r (inside the toroid) . The toroid has a total of N turns . The Amperian loop encloses current Ni. LINES OF . B is constant on the Amperian path. CONSTANT  B ARE  B ds  B  2r   i   iN CIRCLES  0 enc 0 0iN B  inside toroid 2 r • N times the result for a long thin wire • Depends on r • Also same result as for long solenoid N n  (turns/unit length)  B   in 2 r 0 Find B field outside

AMPERIAN LOOP IS Answer B  0 outside A CIRCLE ALONG B Copyright R. Janow – Fall 2013 Find B at point P on z-axis of a dipole (current loop)

• We use the Biot-Savart Law directly   0 i dsrˆ R dB  r  R2  z2 cos  4 r2 r

dB cancels by symmetry (normal to z - axis)

0 i ds cos() dBz  dB||  dB cos()  4 R2  z2

0 i R dBz  ds ds  R df 4 (R2  z2)3 / 2

• Integrate around the current loop on f – the angle at the center of the loop. • The field is perpendicular to r but by symmetry the part of B normal to z-axis cancels around the loop - only the part parallel to the z-axis survives. 2 i is into 0 iR 0 iR 2 Bz  dBz  ds  df page  4 (R2  z2)3 / 2  4 (R2  z2)3 / 2 0

2 as before recall definition of 0i R B(z)  0i Dipole moment 2 2 3 / 2 B(z  0)  2 2(R  z ) 2R Copyright  NiA R. Janow iR – Fall 2013 B field on the axis of a dipole (current loop), continued Far, far away: suppose z >> R  iR2  iR2 Same 1/z3 dependence as for B(z)  0  0 2(R2  z2)3 / 2 2z3 electrostatic dipole  Dipole moment vector is normal to loop (RH Rule).    NiAˆ N  number of turns  1 R2i  |  | above A  area of loop  R2 For any current loop, along z axis with |z| >> R  For charge dipole   0   1 p  B(z)  E(z)  3 3 2 z 20 z

Current loops are the elementary sources of magnetic field: • Creates dipole fields with source strength     • Dipole feels torque to another  in external B field    B  Dipole-dipole   Torque   1 2    r 3 1 2 interaction: dependsCopyright on R. Janow – Fall 2013 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. II. III.

Hint: consider radius, direction, arc angle 0i B  f f in radians 4R Answer: B Copyright R. Janow – Fall 2013 Summary: Lecture 10 Chapter 29 – Magnetic Fields from Currents

Thin wire, asymmetric point  I B  0 [sin( )  sin( )] 4a 1 2 Copyright R. Janow – Fall 2013