Today in Physics 217: Ampère’s Law - Department of ...dmw/phy217/Lectures/Lect_27b.pdf · Today...

6 November 2002 Physics 217, Fall 2002 1

Today in Physics 217: Ampère’s Law

Magnetic field in a solenoid, calculated with the Biot-Savart lawThe divergence and curl of the magnetic fieldAmpère’s lawMagnetic field in a solenoid, calculated with Ampère’s lawSummary of electrostatics and magnetostatics so far

Cd

A

J

π π⋅ = ⋅ =∫ ∫

C Aencl

4 4d d Ic c

B J a


Another Biot-Savart law example: the solenoid

Griffiths problem 5.11: find the magnetic field at point P on the axis of a tightly-wound solenoid (helical coil) consisting of n turns per unit length wrapped around a cylindrical tube of radius a and carrying current I. Express your answer in terms of (it’s easiest that way). Consider the turns to be essentially circular, and use the result of example 5.6. What is the magnetic field on the axis of an infinite solenoid?

θ θ1 2 and

a

I

θ1

θ2 P


Reminder of the result of Example 5.6

Magnetic field a distance zalong the axis of a circular loop with radius R and current I:

( )2

3 22 2

2ˆ I Rc z R

π=

+B z

z

d

d

rr

zdB

sdBzdB

sdB

2 zdB

dBdB

R


The solenoid (continued)

Suppose that n is so large that we can consider the loops in the coil to be displaced infinitesimally; then the number of loops in a length dz is ndz, and

a

I

θ1

θ2 Pz

( )π

=+

2

3 22 2

2ˆ Indz adc z a

B z



Take

so

a

I

θ1

θ2 Pz

( )

θ

θ θθ θ

θ

θ

=

+ = = − = −

⇒ = −

22

2 2

2

tan

tan1 tancos

sin

az

d ad dz dzaz

adz



a

I

θ1

θ2 Pz

( )π π θ θ

θ

πθ θ

= = −

+

= −

2 3

3 2 22 2

2 2 sinˆ ˆsin

2ˆ sin ;

Indz a In addc c az a

In dc

B z z

z

( )θ

θ

π πθ θ θ θ= − = −∫

2

1

2 12 2ˆ ˆsin cos cos .In Ind

c cB z z



For an infinite solenoid, θ θ π= =2 10 and , so

a

I

θ1

θ2 Pz

( )π ππ µ= − = = 0

2 4ˆ ˆ ˆcos0 cos in MKS .In In Inc c

B z z z


The divergence and curl of B

Any vector field is uniquely specified by its divergence and curl. What are the divergence and curl of B? Consider a volume V to contain current I, current density

Denote gradient with respect tothe components of r and r’ by

Now note that

( ) ( )τ

′ ×′= ∫ 2

ˆ1 dc

J rB r

r

r

r’

r

r

P

( )′J r

τ ′d

SV

( )′ :J r

′ and .— — ′ ′= − = −

= − 2

1 1 (because ),

ˆ1and .

r rrr r

rr r

— —

—


The divergence and curl of B (continued)

With these,

This is a useful form for B, which we will use a lot next lecture too (the integral turns out to be the magnetic vector potential, A). Take its divergence:

( ) ( ) ( )

( ) ( )( )

τ τ

τ

′ ′ ′ ′= − × = ×

′′= × ≠

∫ ∫

∫

2ˆ1 1 1

1 remember, .

d dc c

dc

B r J r J r

J rJ f r

r

rV V

V

r

r

—

—

( ) ( )τ

′ ′⋅ = ⋅ × =

∫1 0 .dc

J rB r

Vr

— — —The divergence of any curl is zero, remember.



Integrate this last expression over any volume:

Compare these to the expressions for E in electrostatics, and we see that magnetostatics involves no counterpart of charge: there’s no “magnetic charge.”Now for the curl:

Use Product Rule #10:

( ) τ⋅ = ⋅ =∫ ∫ 0 .d dB r B a—

( ) ( )τ

′′× = × × ∫

1 .dc

J rB r

Vr

— — —

( )× × = ⋅ −∇2 :A A A— — — —



Now use your old friend Product Rule #5,

to write

( ) ( ) ( )

( ) ( )

2

2

1 1

1 1 1 .

d dc c

d dc c

τ τ

τ τ

′ ′ ′ ′× = ⋅ − ∇ ′ ′ ′ ′= ⋅ − ∇

∫ ∫

∫ ∫

J r J rB r

J rJ r

— — —

— —

V V

V V

r r

r r

( ) ( ) ( )⋅ = ⋅ + ⋅ ,f f fA A A— — —

( ) ( ) ( ) ( )τ′ ′ ′ ′ ′⋅ = ⋅ + ⋅ = ⋅

1 1 1d

J rJ r J r J r— — — —

r r r r

=0 (J independent of r)



Also,

so

Use Product Rule #5 again, on the first term:

( )πδ ∇ = ⋅ = ⋅ =

2 32ˆ1 1 4 ,— — — r

rr r r

( ) ( ) ( ) ( )

( ) ( )

πτ δ τ

πτ

′ ′ ′ ′ ′× = ⋅ + −

′ ′ ′= − ⋅ +

∫ ∫

∫V V

V

31 1 4

1 1 4 .

d dc c

dc c

B r J r J r r r

J r J r

— — —

— —

r

r

( ) ( ) ( ) ( )′ ′ ′ ′ ′ ′ ′ ′⋅ = ⋅ − ⋅ = ⋅

1 1J r J rJ r J r— — — —

r r r r

=0 in magnetostatics



So,

But by definition J = 0 on the surface, so the integral vanishes:

This can be put into integral form by choosing an area that some current flows through:

( ) ( ) ( )

( ) ( )

πτ

π

′ ′ ′× = − ⋅ + ′ ′= − ⋅ +

∫

∫

V

S

1 4

1 4 .

dc c

dc c

J rB r J r

J ra J r

— — —

—

r

r

( ) ( )π× =

4 .c

B r J r— Ampère’s Law



Ampère’s law is to magneto-statics what Gauss’ law is to electrostatics, except that one uses an Ampèrean loop to enclose current, instead of a Gaussian surface to enclose charge. The same tricks we learned with Gauss’ law and superposition have analogues in magnetostatics.

Cd

A

J

( ) π

π

× ⋅ = ⋅

⋅ =

∫ ∫

∫A A

Cenclosed

4

4 .

d dc

d Ic

B a J a

B

—


Example: field in an infinite solenoid

The symmetry of the coildictates that the field mustbe along z, and must be a lot stronger inside than out,so if the number of turnsper unit length is n, and thecurrent is I,

∆z

[ ]

π π

π πµ

⋅ = ⋅ =

∆ = ∆ ⇒ = =

∫ ∫C A

enclosed

0

4 4

4 4 ˆ in MKS .

d d Ic c

nIB z In z nIc c

B J a

B z

C A

Same as before!


Maxwell’s equations for electrostatics and magnetostatics

Note the similarities and differences:

πρπ

⋅ = ⋅ =

× = × =

4 040c

Ε B

E B J

— —

— —

π

π

⋅ = ⋅ =

⋅ = ⋅ =

∫ ∫∫ ∫

encl

encl

4 0

40

d Q B d

d B d Ic

E a a

E


Maxwell’s equations for electrostatics and magnetostatics

Note the similarities and differences (MKS):

ρε

µ

⋅ = ⋅ =

× = × =0

0

0

0

Ε B

E B J

— —

— —

ε

µ

⋅ = ⋅ =

⋅ = ⋅ =

∫ ∫

∫ ∫

encl0

0 encl

1 0

0

d Q B d

d B d I

E a a

E

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