magnetostatics - arraytool · 2012-10-04 · gauss law for magnetism ***biot-savart lawampere...

Gauss Law for Magnetism *** Biot-Savart Law Ampere Circuital Law Boundary Conditions Potentials Force Inductance Problems

Magnetostatics

S. R. [email protected]

School of Electronics EngineeringVellore Institute of Technology

October 18, 2012

Magnetostatics EE208, School of Electronics Engineering, VIT

mailto:[email protected]


Outline

1 Gauss Law for Magnetism ***

2 Biot-Savart Law

3 Ampere Circuital Law

4 Boundary Conditions

5 Potentials

6 Force

7 Inductance

8 Problems



Coulomb’s Law for Magnetism - Discrete ***

The force ~Fmir

(ir stands for irrotational) acting on the charge qm located at~r, due to the presence ofthe charge q′m located at ~r′ in an otherwise empty space, is given as

~Fmir(~r) =

qmq′m4πµ0

~r−~r′∥∥∥~r−~r′∥∥∥3 = − qmq′m

4πµ0∇

1∥∥∥~r−~r′∥∥∥ = qm~Hir (1)

where vacuum permeability, µ0 = 4π × 107 Hm−1(SI unit).



Coulomb’s Law for Magnetism - Continuous ***

~dHir

=dq′m

4πµ0

~r−~r′∥∥∥~r−~r′∥∥∥3 =

ρ′m

(~r′)

dv′

4πµ0

~r−~r′∥∥∥~r−~r′∥∥∥3

⇒ ~Hir =1

4πµ0

˚V′

ρ′m

(~r′) ~r−~r′∥∥∥~r−~r′

∥∥∥3 dv′ (2)

= −∇

14πµ0

˚V′

ρ′m

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

(3)



Gauss Law for Magnetism ***

According to Helmholtz’s theorem, a well-behaved vector field is completely known if one knows

its divergence and curl. Taking the divergence of~Bir = µ0~Hir = µ0~Fir

mqm

gives

∇ ·~Bir = −∇ ·

∇ 1

4π

˚V′

ρ′m

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

= − 1

4π

˚V′∇2

1∥∥∥~r−~r′∥∥∥ ρ′m

(~r′)

dv′

=

˚V′

δ(~r−~r′

)ρ′m

(~r′)

dv′

=

˚V′

δ(~r′ −~r

)ρ′m

(~r′)

dv′ = ρ′m (~r) . (4)

Since in reality we do not have magnetic charges, ∇ ·~Bir = 0.



Irrotational Property of Magnetic Field Produced byMagnetic Charges ***

We have already stated in the Coulomb’s law for magnetic charges, that magnetic field produced bymagnetic charges (assuming that they do exist) is always irrotational. Now let’s prove that statementmathematically.

Since ∇× [∇α (~r)] ≡~0 for any R3 scalar field, we immediately find that in magnetostatics

∇× ~Hir = −∇×

∇ 1

4πµ0

˚V′

ρ′m

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

= − 1

4πµ0∇×

∇˚

V′

ρ′m

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

= ~0 (5)

i.e., that ~Hir is an irrotational field.



Outline


2 Biot-Savart Law



5 Potentials

6 Force

7 Inductance

8 Problems



Current

Current Densities - Notations:

Ie; JLe~dl; ~JS

e ds; ~Jve dv

or

I~dl; ~Kds; ~Jdv

Current Density - Definition:

Ie =dQdt

=ρL

e4x4t

= ρLe vx. (6)

In general,

~Jve = ρe~v (7)



Continuity Equation

The current through a closed surface is given as

Ie =

‹~JS

e · ~ds =‹

~Jve · ~ds =

˚ (∇ ·~Jv

e

)dv = − dQ

dt= − d (

˝ρedv)

dt.

From the above equation,

∇ ·~Jve = − ∂ρe

∂t(8)



Biot-Savart Law - Statement

Magnetic field can also be produced by electric currents (Actually, this is the only way we can pro-duce magnetic field in practice). According to Biot-Savart law, magnetic field ~dH

r(r stands for rota-

tional) at position~r generated by a steady differential current element I~dl is given as

~dHr=

I4π

~dl×(~r−~r′

)∥∥∥~r−~r′

∥∥∥3 . (9)



Finite Line Current

Magnetic field due to a small differential current ele-ment at ~r′ = z′ z is given as

~dH =I0δ (x′) δ (y′) dx′dy′

4π

dz′ z×(~r−~r′

)∥∥∥~r−~r′

∥∥∥3

=I0δ (x′) δ (y′) dx′dy′

4π

dz′ z× (xx + yy + (z− z′) z)[x2 + y2 + (z− z′)2

] 32

.

Converting the above equation into cylindrical coordi-nate system gives

~dH =I0δ (x′) δ (y′) dx′dy′

4π

dz′ z× (ρρ + (z− z′) z)[ρ2 + (z− z′)2

] 32

=I0δ (x′) δ (y′) dx′dy′

4π

ρdz′ φ[ρ2 + (z− z′)2

] 32

.



Finite Line CurrentTo get the total magnetic filed, let’s evaluate the volume integral given below:

~Htotal =

˚V′

~dH =

˚V′

I0δ (x′) δ (y′)4π

ρ[ρ2 + (z− z′)2

] 32

dx′dy′dz′ φ

=I0

4π

ˆ z′=b

z′=a

ρ[ρ2 + (z− z′)2

] 32

dz′ φ.

The above integral can be calculated as shown below:

ˆ z′=b

z′=a

ρ[ρ2 + (z− z′)2

] 32

dz′ = ρ

ˆ ξ=z−b

ξ=z−a

1

[ρ2 + ξ2]32(−dξ) , where ξ = z− z′

= ρ

ˆ τ=tan−1(

z−aρ

)τ=tan−1

(z−b

ρ

) 1

[ρ2 + ρ2tan2τ]32

(ρsec2τdτ

), where ξ = ρtanτ

=1ρ

ˆ τ=tan−1(

z−aρ

)τ=tan−1

(z−b

ρ

) cos τdτ =1ρ

z− a√(z− a)2 + ρ2

− z− b√(z− b)2 + ρ2

.



Circular Loop Current

Magnetic field due to a small differential current ele-ment at ~r′ = aρ at the origin is given as

~dH =I0δ (ρ′ − a) δ (z′) dρ′dz′

4π

ρdφ′ φ×(~r−~r′

)∥∥∥~r−~r′

∥∥∥3

=I0δ (ρ′ − a) δ (z′) ρdρ′dφ′dz′

4π

aza3 .

Taking volume integral for the above equation gives

~H =1a2

˚V′

I0δ (ρ′ − a) δ (z′)4π

ρdρ′dφ′dz′ z

=I0

4πa2

[ˆρ′

ρδ (ρ′ − a) dρ′

] [ˆ φ=2π

φ=0dφ′

] [ˆ z′=+∞

z′=−∞δ (z′) dz′

]z

=I0

4πa2 × a× 2πz =I0

2az.



Outline


2 Biot-Savart Law



5 Potentials

6 Force

7 Inductance

8 Problems



Ampere Circuital Law - Statement

Ampere Circuital Law is given (without derivation) as,

˛~H · ~dl = Ienclosed =

¨ (~JS

e · ~ds)=

¨ (~Jv

e · ~ds)

. (10)

The above equation can be combined with Stokes’ theorem to give

∇× ~H =~Jve . (11)



Outline


2 Biot-Savart Law



5 Potentials

6 Force

7 Inductance

8 Problems



Boundary Conditions - Tangential Components

12

34

5

6

7x x x x x + + + + +

Using the Stokes’ Theorem,

(ˆ1+

ˆ3+

ˆ2+

ˆ4

)(~H · ~dl

)=

¨ (∇× ~H

)· ~ds(ˆ

1+

ˆ3

)(~H · ~dl

)=

¨ (~Je · ~ds

), (∵ ∇× ~H =~Je)

⇒(

H1tangential −H2

tangential

)4l = K4l

⇒ H1tangential −H2

tangential = K

⇒B1

tangential

µr1−

B2tangential

µr2= K.



Boundary Conditions - Normal Components

12

34

5

6

7x x x x x + + + + +

Using the divergence theorem (and Gauss law for magnetostatics),

(¨5+

¨6+

¨7

)(~B · ~ds

)= Qm,cylinder

⇒(B2

normal − B1normal

)ds = ρmds

⇒ B2normal − B1

normal= ρm

⇒ µr2H2normal − µr1H1

normal=ρm.



Outline


2 Biot-Savart Law



5 Potentials

6 Force

7 Inductance

8 Problems



Magnetic Field - Sources

Magnetic field can be produced either from magnetic charges (ρm) or from electric current (I~dl).Magnetic field produced by ρm is always irrotational, whereas magnetic field produced by I~dl isrotational.

~Htotal = ~Hirrotational + ~Hrotational



Gauss Law - Revisit

Gauss Law:

∇ ·~B = ρm

⇒ ∇ ·(~Birrotational +~Brotational

)= ρm + 0

It can be shown that

∇ ·~Birrotational = ρm, and

∇ ·~Brotational = 0 (12)



Ampere Law - Revisit

Ampere Law:

∇× ~H =~Je

∇×(~Hirrotational + ~Hrotational

)=~0 +~Je

It can be shown that

∇× ~Hrotational =~Je, and

∇× ~Hirrotational =~0 (13)



Scalar and Vector Magnetic Potentials

From (12) , (13), and the above diagram, we can introduce two potentials as shown below:

∇ ·~Brotational = 0 ⇒ ~Brotational = ∇×~A (14)

∇× ~Hirrotational =~0 ⇒ ~Hirrotational = −∇Vm (15)

In the above equations, ~A and Vm are called, vector and scalar magnetic potentials, respectively.

Since we don not have magnetic charges, we will concentrate only on vector magnetic potential.



Vector Magnetic PotentialMathematical expression for vector magnetic potential ~A can be derived from Biot-Savart Law asshown below: (The below derivation is not in your syllabus)

~H =1

4π

˚~Je

(~r′)×

(~r−~r′

)∥∥∥~r−~r′

∥∥∥3 dv′

=−14π

˚~Je

(~r′)×∇

1∥∥∥~r−~r′∥∥∥ dv′

=−14π

˚ 1∥∥∥~r−~r′∥∥∥(∇×~Je

(~r′))−∇×

~Je

(~r′)

∥∥∥~r−~r′∥∥∥ dv′ ,∵ ∇× (wA) = w∇×A−A×∇w

=−14π

˚ ~0−∇× ~Je

(~r′)

∥∥∥~r−~r′∥∥∥ dv′ = ∇×

14π

˚ ~Je

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

. (16)

Comparing (16) and (14), we get (you need to remember the below formula)

~A =µ0

4π

˚ ~Je

(~r′)

∥∥∥~r−~r′∥∥∥ dv′ (17)



Outline


2 Biot-Savart Law



5 Potentials

6 Force

7 Inductance

8 Problems



Lorentz and Laplace Forces

Lorentz Force:According to the Lorentz force equation, force on a moving charge is given as,

~F =~Fe +~Fm = Q(~E +~u×~B

)(18)

Laplace Force:Similarly, force acting on a differential current element due to an incident magnetic field ~B is givenas

~Fm = I~dl×~B (19)



Ampere Force Law



Outline


2 Biot-Savart Law



5 Potentials

6 Force

7 Inductance

8 Problems



Inductance



Mutual Inductance



Magnetic Energy Density



Inductance - Examples



Outline


2 Biot-Savart Law



5 Potentials

6 Force

7 Inductance

8 Problems



Biot - Savart Law

1 If~I = 14z A, then the magnetic field intensities ~H atPCart (2,−4, 4) and PCart

(√20, 0, 4

)are given as ____.

2 Find the magnetic field intensity at the center of a circularconductor carrying a current of 10 A with a radius of 2m withorigin as center.

3 The magnetic field intensity at the center of a square conductorof 5m each side carrying a current of 5 A with origin as centerlying in z = 0 plane is ____.

4 The flux φ crossing the plane surface defined by0.5m ≤ ρ ≤ 2.5m and 0m ≤ z ≤ 2m if~B is given by~B = 2

ρ affi.

5 Given~Je = 103 sin θar A/m2 in spherical coordinates, then thecurrent crossing the spherical shell r = 0.02m is ____.



Ampere Circuital Law

Toroid



Boundary Conditions

1 Plane y = 1 carries current ~K = 50z mA/m. Then find the magnetic field intensities at(0, 0, 0) and (1, 5,−3), respectively. (25x,-25x)

2 The plane z = 0 marks the boundary between free space and a dielectric medium with a

dielectric constant of 40. The~E field next to the interface in free space is~E = 13x + 40y + 50z V/m. Then find the~E field on the other side of the interface.

(13x+40y+5/4z)



Potentials



Lorentz Force

force acceleration ... etc



Faraday’s Law


magnetostatics - arraytool · 2012-10-04 · gauss law for magnetism ***biot-savart lawampere...

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