Magnetic Field

 

Biot Savart Law

Biot Savart law states that "at any point P the magnitude of magnetic field intensity produced by the differential element is directly proportional to the product of current, magnitude of differential length and sine of angle lying between filament and line connecting filament to point P; and inversely proportional to the square of distance from differential element to point P.

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Derivation

Consider a filament carrying current I.
According to Biot Savart law;

     dH = ( I * dl * sinθ ) / ( 4 π * R ^ 2 )
In vector form;
     dH vector = I (dL × a(R)) / ( 4 π * R ^ 2 )

 

The unit of magnetic field intensity H is A/m.

 

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Magnetic Field Intensity due to a filament carrying current

Consider an infinitely long filament carrying current I placed on Z-axis. We have to find vector H at point P due to this filament using Biot Savart law.

From figure:
r = y a(y) = p a(p)
r' = z a(z)
R = r - r' = pa(p) - za(z)

Also,
dl = dz a(z)

From Biot Savart law;

dH = I (dL × a(R)) / ( 4 π * R ^ 2 )
      = I (dz a(z) × (pa(p) - za(z))) / ( 4 π * (z ^ 2 + p ^ 2) ^ (3/2) )
      = I p dz a(ϕ) / ( 4 π * (z ^ 2 + p ^ 2) ^ (3/2) )

 

The total magnetic field intensity is given by:

H = ∫(-∞, ∞) [ I p dz a(ϕ) / ( 4 π * (z ^ 2 + p ^ 2) ^ (3/2) ) ]
    = I p a(ϕ) / ( 4 π)  *  { ∫(-∞, ∞) [ dz / (z ^ 2 + p ^ 2) ^ (3/2) ]}
    = I p a(ϕ) / ( 4 π)  * 2 / p^2
    = I / 2πp * a(ϕ)

The direction of H is circumferential

 

In case of finite filament
H = I / 4πp * (sinα2 - sinα1) * a(ϕ)

The angle which lies below point P is taken negative.

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Ampere's Circuital Law

Surface Current Density

- It is defined as the current flowing per unit width.
- For uniform current, I = kb
- For non-uniform current, I = ∫ kdN
where, dN is incremental length along filament carrying current I

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Ampere's Law

Ampere's circuital law states that "the line integral of H about any closed path is exactly equal to the direct current enclosed by that path."

So, ∮H.dl = I(enclosed)

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H due to infinite filament carrying current I

Here, H = H(ϕ) . a(ϕ)

Consider circular amperian path;

     ∮H.dl = I(enclosed)
or, H(ϕ) ∮a(ϕ).ρ d(ϕ) a(ϕ) = I
or, H(ϕ) ρ ∮d(ϕ) = I
or, H(ϕ) ρ [ ϕ](0 to 2π) = I
or H(ϕ) = I / 2πρ

 

In vector form;
H = I / 2πρ * a(ϕ)

 

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H in coaxial cable

Consider an infinite long coaxial cable carrying uniform current I in inner conductor and -I in outer conductor.

From symmetry, choose a circle of radius ρ as amperian closed path.

 

1. For a<ρ      ∮H.dl = I(enclosed)
or, H(ϕ) ∮a(ϕ).ρ d(ϕ) a(ϕ) = I
or, H(ϕ) ρ 2π = I
or H(ϕ) = I / 2πρ

 

2. For ρ ∮H.dl = I(enclosed) or, H(ϕ) . 2πρ = (I / πa^2) * πρ^2 or, H(ϕ) = Iρ / 2πa^2

3. For b<ρ ∮H.dl = I(enclosed) or, H(ϕ) . 2πρ = I * [(c^2 - ρ^2) / (c^2 - b^2)] or, H(ϕ) = ( I / 2πρ ) * [(c^2 - ρ^2) / (c^2 - b^2)]

4. For ρ>c; I(enclosed) = 0 So, H(ϕ) = 0

 

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Magnetic Flux Density

- The magnetic flux through any surface is given by:
ϕ = ∫s [ B.dS ]

- The unit of magnetic flux is Weber (Wb).

- Magnetic flux is circular in nature. We can separate positive and negative charges but monopoles can not be created.

 

- From Maxwell's equation;
           ∇ . B = 0

 

Integrating both sides w.r.t volume:
∫v [ (∇ . B) dv ] = 0

Applying divergence theorem;
∮s [ B.dS ] = 0
Hence, for any closed surface, ϕ = 0.

Magnetic flux density is given by:
B = μo H
where, μo = 4π * 10 ^ -7 H/m

 

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Physical Significance of Curl; Stoke's Theorem

Curl

 

Consider an incremental closed path of sides Δy and Δz. Let some current produces H at the centre of this rectangle P(xo, yo, zo).

 

So, H(o) = H(xo) a(x) + H(yo) a(y) + H(zo) a(z)

The closed line integral of H about this path is equal to the sum of four values of H.ΔL on each side. Choose a direction as a-b-c-d-a, which corresponds the current in u(x) direction.

So, (H.ΔL)a-b = H(y, a-b).Δy

The value of H(y, a-b) in reference to H(yo) at center is:
H(y, a-b) = H(yo) - δH(y)/δz (1/2 Δz)

Hence, (H.ΔL)a-b = [ H(yo) - δH(y)/δz (1/2 Δz) ] . Δy

Similarly,
(H.ΔL)b-c = [ H(zo) - δH(z)/δy (1/2 Δy) ] . Δz
(H.ΔL)c-d = - [ H(yo) - δH(y)/δz (1/2 Δz) ] . Δy
(H.ΔL)b-c = - [ H(zo) - δH(z)/δy (1/2 Δy) ] . Δz

Now,
∮H.dl = - δH(y)/δz . ΔyΔz + δH(z)/δy . ΔyΔz = [δH(z)/δy - δH(y)/δz] . ΔyΔz

By Ampere's circuital law;
lim(ΔyΔz --> 0) [∮H.dl / ΔyΔz ] = [δH(z)/δy - δH(y)/δz] = ΔI / ΔyΔz = J(x) .......eqn1

Similarly,
lim(ΔzΔx --> 0) [∮H.dl / ΔzΔx ] = [δH(x)/δz - δH(z)/δx] = ΔI / ΔzΔx = J(y) .......eqn2
lim(ΔxΔy --> 0) [∮H.dl / ΔxΔy ] = [δH(y)/δx - δH(x)/δy] = ΔI / ΔxΔy = J(z) .......eqn3

Hence, the above equations gives the component of curl which is normal to the area enclosed by closed path.
So, [curl A](N) = lim(ΔS(n) ---> 0) [∮A.dl / ΔS(n)]

Therefore,
Curl H = [δH(z)/δy - δH(y)/δz] a(x) + [δH(x)/δz - δH(z)/δx] a(y) + [δH(y)/δx - δH(x)/δy] a(z)

Generally,
curl H = ∇ × H = Magnitude of matrix given by

ax ay az
δ/δx δ/δy δ/δz
Hx Hy Hz

According to Maxwell;
∇ × H = J

 

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Physical Interpretation of Curl

- ∮H.dl and ∮E.dl explains the circulation of E and H respectively.
- Curl gives the circulation per unit area.
- Curl provides the maximum circulation of field at any point per unit area and gives the direction of maximum circulation.
- If there is no circulation, it means curl is equal to zero.

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Stoke's Theorem

According to Stoke's Theorem;
∮H.dl = ∫s [ (∇ × H) . dS ]

The closed line integral of H is equal to the open surface integral of curl of H.

Consider surface S which is divided into incremental surfaces ΔS = Δs a(n).

From the definition of curl applied to one of the incremental surface ΔS;

    ∮H.dl(direction ΔS) / Δs = (∇ × H) (direction N) ................eqn 1
         where, N denotes component of curl normal to ΔS
                      dl(direction ΔS) denotes perimeter enclosing ΔS

 

So, eqn 1 can be written as:
∮H.dl(direction ΔS) / Δs = (∇ × H) . a(N)
∮H.dl(direction ΔS) = (∇ × H) . Δs.a(N)
∮H.dl(direction ΔS) = (∇ × H) . ΔS

Applying to all incremental surfaces;
∮H.dl = ∫s [(∇ × H) . dS]

This proves Stoke's theorem

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Divergence of curl of vector

- The divergence of a curl is equals to zero.

Let for any vector A, ∇ . (∇ × A) = T
where, T being any scalar.

On integrating w.r.t volume;
∫v [∇ . (∇ × A) dv] = ∫v T dv

Applying divergence theorem;
∮s [(∇ × A) dS] = ∫v T dv

It is clear that the close path enclosing a surface is 0.
So, ∮s [(∇ × A) dS] = 0

i.e ∫v T dv = 0
As, ∫v [dv] is not equal to 0, T = 0.

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Scalar and Magnetic Vector Potential

Scalar Vector Potential

For a magnetic field, consider Vm to be the scalar magnetic potential.
So, H = -∇.Vm -----1

From Maxwell's equation;
∇ × H = J -------2

From 1 and 2;
∇ × (-∇.Vm) = J
So, J = 0

This shows that scalar magnetic potential is valid only for those regions where J = 0.

From Maxwell's equation,
∇ . B = 0
∇ . (μH) = 0
μ ( ∇ . H ) = 0
∇ . ( - ∇ . Vm) = 0
∇^2 Vm = 0

This shows that scalar magnetic potential satisfies Laplacian equation.

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Scalar Magnetic Potential at certain point is not unique

Consider a C.S.A of coaxial line. We know, in the region a<δ

H = I / 2πρ . a(ϕ) -------1
H = - ∇.Vm --------2

From 1 and 2;
∇.Vm = - I / 2πρ . a(ϕ)
(1/ρ) . (δVm / δϕ) . a(ϕ) = - I / 2πρ . a(ϕ)
δVm = - I / 2π * δϕ

On integrating;
Vm = - I / 2π * ϕ [Assuming constant to be zero]

For ϕ = 0; Vm = 0
For ϕ = 2π; Vm = -I
For ϕ = 4π; Vm = -2I
For ϕ = π/4; Vm = -I/8
For ϕ = π/4 + 2π; Vm = -(1-1/6)I
For ϕ = π/4 + 4π; Vm = -(2-1/8)I

So, Vm = I (n - 1/8) where, n = 0, 1, -1, 2, -2, .............

This shows that Vm is not unique at any specific point.

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Vector Magnetic Potential

We know,
∇ . B = 0

From vector identity, B = ∇ × A
where, A is a vector magnitude potential.

So, ∇ . (∇ × A) = 0

Now,
B = ∇ × A
μo H = ∇ × A
H = (∇ / μo) × A
∇ × H = 1/μo ( ∇ × ∇ × A)

Hence, vector magnetic potential is given by:
A = ∫ [ (μo I dl) / (4 π R) ]

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Magnetic Boundary Conditions

Consider the boundary condition as shown in given figure:

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According to Gauss law;
       ∮s [ B.dS ] = 0
or,   B(N1) . ΔS - B(N2) . ΔS = 0
or,   B(N1) = B(N2) ..................................eqn 1

 

Also,
H(N2) = μ1 / μ2 * H(N1)

M(N2) = χ(m2) * μ1 / μ2 * H(N1) = {χ(m2) / χ(m1)} * {μ1 / μ2} * M(N1)

M(N1) = χ(m1) * H(N1)

 

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According to Ampere's Law:
        ∮ [ H.dl ] = I
or,    H(t1).ΔL - H(t2).ΔL = I
or,    H(t1) - H(t2) = I

 

Also,
(H1 - H2) × a(N12) = K

B(t1) / μ1 - B(t2) / μ2 = k

 

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Numerical Problems of Magnetic Field

1. A current filament carrying 15A in a(z) direction lies along entire Z-axis. Find H at point A(√20, 0, 4).

 

At point A:

 

ρ = √[x^2 + y^2] = √20
ϕ = tan-1 (y/x) = 0°

H = I / 2πρ * a(ϕ) = 0.534 a(ϕ)

Let H = Hx a(x) + Hy a(y) + Hz a(z)

Then,
Hx = H.a(x) = 0.534 a(ϕ) . a(x) = -0.534 sinϕ = 0
Hy = H.a(y) = 0.534 a(ϕ) . a(y) = 0.534 cosϕ = 0.534
Hz = H.a(z) = 0.534 a(ϕ) . a(z) = 0

Hence, H = 0.534 a(y)

 

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2. Determine H at point P(0.4, 0.3, 0) in field of 8A filamentary current directed inward from infinity to origin on positive X-axis and then outward to infinity along Y-axis.

 

Angle made by line joining P and Y-axis perpendicularly and line joining P and origin (α1y) = -tan-1(0.3/0.4) = -36.9
Angle made by line joining P and X-axis perpendicularly and line joining P and origin (α2x) = -tan-1(0.4/0.3) = 53.1
Angle made by line joining P and Y-axis perpendicularly and line joining P and X-axis perpendicularly is given by: (α1x) = -90  and (α2y) = 90

 

For current in X-axis, the radial distance ρ is measured from X-axis; ρ = 0.3
H(x) = [8 / (4π*0.3)] * [sin53.1 + 1] [-a(z)] = -12/π a(z)

For current in Y-axis, the radial distance ρ is measured from Y-axis; ρ = 0.4
H(y) = [8 / (4π*0.4)] * [1 + sin36.9] [-a(z)] = -8/π a(z)

Hence,
H = H(x) + H(y) = -20/π a(z)

 

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3. Calculate vector current density in spherical coordinates at (2, 30°, 20°) if H = 1/sinθ a(θ).

 

The vector current density is given by:
     J = ∇ × H

 

Now,
∇ × H is given by magnitude of matrix given by:
1/(r^2 sinθ) a(r) 1/rsinθ a(θ) 1/r a(ϕ)
δ/δr δ/δθ δ/δϕ
Hr r Hθ rsinθ Hϕ

We have, Hr = 0 = Hϕ and Hθ = 1/sinθ

So,
∇ × H = 1/(r^2 sinθ) a(r) [δ(r/sinθ) / δϕ] + 1/r a(ϕ) [δ(r/sinθ) / δr]
= [ 1 / (r sinθ) ] a(ϕ)

At (2, 30°, 20°),
J = ∇ × H = a(ϕ)

 

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4. Calculate total current passing through path around region 2<=x<=5, -1<=y<=1 and z = 0 where H = 6xy a(x) - 3y^2 a(y) A/m.

 

∇ × H is given by magnitude of matrix given by:
                                       ax           ay           az
                                       δ/δx       δ/δy       δ/δz
                                       6xy          -3y^2        0

 

So, ∇ × H = -6x a(z)

Now,
∫s [ (∇ × H).dS ] = ∫(x=2 to 5) ∫(y= -1 to 1) [-6x a(z) . a(z) ] dx dy
= -126 Ampere

Using Stoke's theorem and Ampere's circuital law:
∮H.dl = ∫s [ (∇ × H) . dS ]
∮H.dl = I(enclosed)

We get, I(enclosed) = -126 A

 

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5. Find H at origin if K = 2 a(z) A/m flows in plane x = -2.

 

We have,
    K = 2 a(z)

 

The plane is:
x = -2

Since, the point (0, 0, 0) is on right of the given plane;
H = 1/2 * [K × a(x)]
= 1/2 * [2 a(z) × a(x)]
= a(y)

 

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6. A square loop with A(0, -2, 2), B(0,2,2), C(0,2,-2) and D(0,-2,-2) is carrying current of 20A along ABCDA. Find H at center O(0,0,0).

 

The magnetic field at O due to AB is given by:
     H(due to AB) = I / 4πρ * [sinα2 - sinα1] a(ϕ) 
where. sinα2 = sinα1 = 45°

 

H(due to AB) = 20 / 4π*2 [sin45 + sin45] (-a(x)) = -1.125 a(x) A/m

Due to symmetry,
H = 4 H(due to AB) = - 4.501 a(x) A/m