PHY2049: Chapter 29 1

Chapter 29: Magnetic Fields Due to Currents

PHY2049: Chapter 29 2

Unlike the law of static electricity, comes in two pieces

Piece 1: Effect of B field on moving charge

Piece 2: B field produced by current

Biot-Savart Law Ampere’s LawProof of equivalence not in the book (Requires vector calculus and relies on the absence of magnetic monopoles)

Reminds you of similar equivalence between Coulomb’s law and Gauss’ law

Equivalent!

Law of Magnetism

BvqFrr

×= (Chapt. 28)

PHY2049: Chapter 29 3

Creating Magnetic FieldsSources of magnetic fields

Electric current (moving charges)Atomic orbits of electrons (angular momentum L > 0 only)Internal “spin” of elementary particles (mostly electrons)

Magnetic field produced by current is fundamentalHow about field produced by a bar magnet? Bar magnet ← magnetic ions ← orbital motion and spin of electrons in them ← they are microscopic currents

Three examples studied hereLong wireWire loopSolenoid

PHY2049: Chapter 29 4

B Field Around Very Long WireField around wire is circular, intensity falls with

distanceDirection given by RHR #2 (compass follows field lines)

02iBr

µπ

=

70 4 10µ π −= ×

Right Hand Rule #2

Derived from Ampere’s law

PHY2049: Chapter 29 5

(continued)Why does µ0 have such a simple value?

Magnetism is inseparable from electricity. This allows the units in electricity and magnetism (in particular coulomb and tesla) to be chosen so that only one constant, ε0, has a non-trivial value.

This example illustrates important general property of magnetic fields:

Magnetic field lines have no beginning/end, unlike electricfield lines.

PHY2049: Chapter 29 6

Long Wire B Field Example

I = 500 A toward observer. Find B vs rRHR #2 ⇒ field is counterclockwise

r = 1 mm B = 0.10 T = 1000 gaussr = 1 cm (~0.4”) B = 0.010 T = 100 gaussr = 10 cm (~4”) B = 0.001 T = 10 gauss

( )70

4 10 500 0.00012 2iBr r r

πµπ π

−×= = =

PHY2049: Chapter 29 7

Charged Particle Moving Near Wire

Wire carries current of 400 A upwardsProton moving at v = 5 × 106 m/s downwards, 4 mm from wireFind magnitude and direction of force on proton

SolutionDirection of force is to left, away from wireMagnitude of force at r = 4 mm

Iv

02IF evB evr

µπ

= =

( )( )7

19 6 2 10 4001.6 10 5 100.004

F−

− × ×= × ×

141.6 10 NF −= ×

PHY2049: Chapter 29 8

Ampere’s Law First (Biot-Savart law later)

Take arbitrary closed path around set of currentsLet ienc be total enclosed current (signs +/– according to RHR #2)Let B be magnetic field, and ds be differential length along path

Direction of field due to each current element obeys RHR #2

Only currents inside path count!5 currents inside path (included)1 outside path (not included)This does not mean that current outside path does not contribute to B (note similarity to Gauss’ law)

Not includedin iencenc0iµd =⋅∫ sB

PHY2049: Chapter 29 9

Let’s try this for long wire. Find B at distance at point P

According to RHR #2, B field has only azimuthalcomponent, no radial componentDraw circular path passing through P (radius r)From symmetry, strength of B must be constant along path

An easy derivation

Ampere’s Law For Straight Wire

r

P( ) 0

0

2

2

d B r i

iBr

π µ

µπ

⋅ = =

=

∫ B s∫ ⋅ sB d

PHY2049: Chapter 29 10

20

2 RπirµB =

Ampere’s Law: More Application

Find B vs r inside long wire, assuming uniform current

Wire radius R, total current iDraw circular path of radius r

Key fact: enclosed current ∝ areaInside

Outside: (derived in previous slide)

2

2

enc Rrii = r

R

02iBR

µπ

= On surface

( ) enc02 iµrπB =

rπiµB

20=

PHY2049: Chapter 29 11

Question 10Figure shows the magnitude of B field inside and

outside four long wires. Current is uniformly distributed in each wire. Which wire carries the largest current?

(a) 1(b) 1 and 2(c) 1 and 3(d) Insufficient info

In which wire is the current density the highest?

(a) 1(b) 1 and 2(c) 1 and 3(d) Insufficient info

r

B1

3

4

2

PHY2049: Chapter 29 12

Force Between Two Parallel CurrentsForce on I2 from I1

RHR ⇒ Force towards I1

Force on I1 from I2

Must be the same and towards I2

Why? Newton’s third lawOr view from behind the screen.(I1 is now on left, and I2 now on right.)

Magnetic forces attract two parallel currents

I1I2

0 1 0 1 22 2 1 2 2 2

I I IF I B L I L Lr r

µ µπ π

= = =

I1I2

PHY2049: Chapter 29 13

Force Between Two Anti-Parallel CurrentsForce on I2 from I1

RHR ⇒ Force away from I1

Force on I1 from I2

Must be the same and away from I2

Magnetic forces repel two antiparallel currentsI1I2

I1I2

0 1 0 1 22 2 1 2 2 2

I I IF I B L I L Lr r

µ µπ π

= = =

PHY2049: Chapter 29 14

Parallel Currents (cont.)Look at them edge on to see B fields more clearly

Antiparallel: repel

FF

Parallel: attract

F F

B

BB

B

2 1

2

2

2

1

11

PHY2049: Chapter 29 15

Question 6

OUT IN OUT IN OUT

OUT OUT IN IN IN

OUT IN OUT OUT IN

IN OUT IN IN IN

Long wires, carrying equal currents, are parallel to each other and equally spaced. In which arrangement is the net force on the central wire the largest?

(a) a(b) b(c) c(d) d (e) Insufficient info

Wires are of equal lengths.(a) a(b) b(c) c(d) d