Lecture 15 Biot-Savart Law, Ampere’s Law and Magnetic Dipole

8/2/2019 Lecture 15 Biot-Savart Law, Ampere’s Law and Magnetic Dipole

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EECS 117 Lecture 15: Biot-Savart Law,

Ampere’s Law and Magnetic Dipole

University of California, Berkeley



Biot Savart Law and Ampere’s Law

In the last lecture, we have shown that the magnetic forceexerted on a small segment of wire flowing a current I withlength d l is equal to

where B is the magnetic flux density, and . The H is

called the and µ is the permeability of the medium, which is themedium properties just like ε.

The magnetic field intensity H induced by a current carrying wiresegment has been shown to be

Bl Id dF m ×r

H B µ

dvr r

r r r J

r r

r r l d r I r H d 33

'4

)'()'(

'4

)'()'()(

rr

rrr

rr

rr

r

−

−=−

−=π π



Biot Savart Law and Ampere’s Law (cont.)

is a vector pointing towards the point of interest from theorigin, and is a vector towards the wire segment. The secondexpression is for a non-uniform current distribution in the

magnetic field generating system (like a wire).

Ampere’s circuital law states that the magnetic field intensity

integrated along a closed path is equal to the current flowingthrough the surface enclosed by the path.

r r

'r r

I l d H =∫rr



Off-axis Magnetic Field of a Straight Wire

Consider a segment of a straight wire of length 2a. Using Biot-Savart Law, themagnetic flux density due to a small segmentdl is given by

The B vector is pointing into the paper.

))'((4sin

ˆ

'4)'(ˆ)',,(

220

30

z z r dz I

r r r r z Id z r Bd

−

−−=

π α µ

φ

π µ φ φ

rr

rrr

(r,φ,z’)

a

-a

z'r r

rr

−

y

r r

I

α

−−

−++

=−−

=

−−

−

−∫

2222

02/1222

0

2/3220

220

)'(

'

)'(

'4))'((4

)'(

))'((4ˆ

))'((4sinˆ)',,(

a z r

a z

a z r

a z r

I z z r r

z z Ir

z z r Irdz

z z r dz I z r B

a

a

a

a

a

a

π µ

π µ

π µ φ

π α µ φ φ



On-axis Magnetic Field of a Straight Wire

If the point of interest is at the midpoint of the wire, z’ = 0, and

22

0

2

ˆ

ar r

Ia B

+

=

π

µ φ



On-axis Magnetic Field of a Circular Loop

Consider now the magneticfield along the axis of a circular loop wire.

The green segment induces amagnetic field in green.

Because of circular symmetry,the segment (red) on theopposite side create a field (red)of equal magnitude. Theresultant field is along the z-axis.

y

z

x

(0,0,z’)

I

a

Rr



On-axis Magnetic Field of a Circular Loop

2/322

20

2

02/322

20

2/322

20

2/3220

20

)(2ˆ

)(4ˆ

)(4)(4

4

ˆ)',0,0(

z aa I k d

z aa I k B

z ad a I

z aadl I

dB

R

Rl Id z Bd

z

++

++

×=

∫ µ φ π

µ

φ π

µ π

µ

π

µ

π r

r

r

k a I

B ˆ20µ

=r

At the center of the loop,



On-axis Magnetic Field of a Solenoid

z

-l l

Using the result above, we can calculate the on-axismagnetic field inside the solenoid

Consider a solenoid with n number of windings per meter. The length of the solenoid is 2 l.

22

0

222

20

2/322

20

2/322

20

2)(2

)(2

l a

NI

z aa

z nIa

z a

dz nIa B

z a

dz anI dB

l

l

l

l

z

z

+

=

+

=+

+

−

∫ µ µ µ

µ



On-axis Magnetic Field of a Solenoid (cont.)

where N = 2n l, the total number of windings in thesolenoid.

For an infinitely long solenoid, nI B z 0µ



Ampere’s Circuital Law Version of a Solenoid

A B

CD

C’D’

l B

r

The crosssection of thesolenoid.

We can obtain the same information, and even learn more byusing the Ampere’s circuital law.

The line integral of the B can be broken into the segment parts

∫∫ + DACD BC AB

dl Br



Solenoid (Ampere’s Circuital Law)

Based on the result we obtained in the previous calculation, insidethe solenoid is only in the z direction. The geometry implies that thiswould be true also for the magnetic field outside the solenoid. So,

If we consider the path ABC’D’, the path integrals along BC’ andD’A are zero because of the above reason. Because there is not current

flowing through the area enclosed by the closed path,

This result holds if the hypothetical box ABC’D’ is drawn such thatthe segment C’D’ is at infinity. This implies that the magnetic field isuniform everywhere outside the solenoid. This contradicts our intuitionof a field falling down to zero at infinity and also experimental

observation.

∫∫ −='' DC AB

B

0== ∫∫ DA BC

r



Solenoid (Ampere’s Circuital Law)

The only solution to this contradiction is that the magnetic fieldoutside of solenoid is zero. This coincides with experimental results.

The result is:

If the segment CD is far away from the ends of the solenoid, B isuniform along the segment. Thus,

This result is the same as the one for the infinite solenoid.

If we move the segment CD up and down, since the current enclosedremains the same, the path integral along CD is the same. So themagnetic field is uniform inside the solenoid. For an infinite solenoidor the region where end effects are negligible, the field everywhereinside the solenoid is equal to the above result.

nlI l d Bl d BCD

0µ ==

∫∫rrrr

nI BnlI Bl l d BCD 00 µ µ ===∫

rr



Vector Potential

Since there is no magnetic monopole , can beexpressed as , the curl of vector potential.

For the two different expressions of B, we have

0=Br

Br

Av

×

∫∫ −−×=

−−×= dv

r r

r r r J

r r

r r l d r I r B 33

'

)'()'(4'

)'()'(4

)(rr

rrr

rr

rr

r

π µ

π µ

∫∫ −=

−= dv

r r

r J

r r

l d r I r A 22

'

)'(4'

)'(4

)(rr

r

rr

r

π µ

π µ



Magnetic Dipole

x

z

(r,0, θ)

a

θr

φ’ψ

Top view

x

R

For a current loop, the magnetic field off axis can be calculated by finding vector potential.



Magnetic Dipole

The vector potential due to the green segment has a direction of the green arrow. Because of symmetry, the potential due to thered segment almost cancel the green vector potential and theresultant potential due to these two segments has a directionnormal to the xz plane, or in general in direction. Without lossof generality, we place the point of interest in the xz plane.

The φcomponent of A due to a segment is equal to

The total is found as the integral around the loop

φ ̂

R Idl

dAπ

φ µ φ 4

'cos'=

∫=π

φ

φ φ π

µ π

φ µ 2

0

''cos44

'cos' R

d Ia R

Idl A



Magnetic Dipole

where

For r >> a

or

The total vector potential is

are then equal to

2cos222222

−ar raar R ψ 'cossin φ θ ra

'cossin21 φ θ r a

r R − + 'cossin111

φ θ r a

r R

( )222

0

2

4sin''cossin'cos

4 r a I d

r a Ia A

π θ π µ φ φ θ φ

π µ

π

φ ∫ =+

Br

( ) ( )θ θ θ θ

φ φ φ φ ˆsinsin

ˆˆ∂∂−∂∂= rA

r r A

r r A B

r

( )[ ]θ θ θ π

π µ sinˆcosˆ

2 3

2

+r r

a I



Magnetic Dipole

The magnitude of the magnetic dipole moment, the counterpartof electric dipole, is defined as

The dipole direction is pointing up from the loop for the counter-clockwise I direction (as in this case), and vice versa.

The vector potential can then be expressed as

2a I m π

r m A1

40 r

r

π µ

Lecture 15 Biot-Savart Law, Ampere’s Law and Magnetic Dipole

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