There are two methods of magnetic induction B calculation. The first method is using the Biot-Savart-Laplace and induction superposition principle. The second method is using the B vector circulation theorem, which will be explained below.

It is very easy to use this theorem with cylindrical, sphere or flat symmetry of a conductor. Vector circulation by the enclosed loop is

A(B,dr)

Figure 28. (B,dris a projection of B on the vector dr, which is integral algebraic sum of the projections of B on the dr vector, when the loop A is a set of fragments. It is important to note that we cannot derive the loop to the electrons, because an electron is an accelerating particle. However, magnetic field devices in our approximation are characterised by a velocity component. Thus, vector B circulation theorem states the following:

Magnetic field B circulation by the enclosed loop A, is proportional to the algebraic sum of currents going through this loop

A(B,dr)=μ0iIi
Figure 28. Depiction of vector B circulation
Figure 28. Depiction of vector B circulation

If there is a distributed current flowing through the surface Ω surrounded by the closed loop or path A, then

iIi

should be replaced by

Ω jdS

where j is a density of a current flow through the surface Ω enclosed by the loop A.

If we generalise the circulation statements for simple cases, it will lead us to the most complex case.

  1. Considering a wire with electric current, we choose the path A perpendicular to the conductor – this path will correlate to the one of magnetic induction lines. Figure 29. Then
    (B,dr)=B(r)dr=B(r)2πr By the Biot-Savart-Laplace Law B(r) this conductor isμ0I2πr then(B,dr)=μ0I2πr*2πr=μ0I This result corresponds to the theorem we stated before.
Figure 29. Circulation theorem depiction for a conductor, perpendicular to the chosen contour
Figure 29. Circulation theorem depiction for a conductor, perpendicular to the chosen contour 

2. 2. Let’s consider the case with the same straight wire, but taking a closed path with a random form. Then

(B,dr)=Brdr=B*r*dα

The magnetic induction for the wire is

B(r)=μ0I2πr

then

(B,dr)=μ0I2πdα=μ0I2πdα=μ0I
Figure 30. Circulation theorem depiction for a conductor inside the chosen contour
Figure 30. Circulation theorem depiction for a conductor inside the chosen contour

3. Let’s consider the case when the conductor is outside of the path A. Figure 31. The path with be divided into two parts, then the result will be evident:

(B,dr)=μ0I2πr(12dα+21dα)=0
Figure 31. Circulation theorem depiction for a conductor outside the chosen contour
Figure 31. Circulation theorem depiction for a conductor outside the chosen contour

4. If the path A is oriented with some angle to the conductor and has a random form, we can also divide projections of B on the dr – to parallel and perpendicular components:

(B,dr)=Bdr+Bdr=μ0I

(perpendicular component – result related to previous case).

5. In the case of several conductors crossing the loop, we can generalise the result above and make an algebraic sum of the magnetic induction circulation integrals. This generalisation will lead us to the result

μ0jsdS

Let’s resolve some examples using the theorem of magnetic field circulation.

Example 1. The conductor with the radius R is characterised by the uniform current density j, μio=1. Determine the magnetic field B(r) along the axis of the wire

For radius-vector r coordinates are (0,0,z), for radius vector r1 coordinates are (Rcosα,Rsinα,0). Then for dr1 (-Rsinαdα,Rcosαdα,0) and r- r1=(-Rcosα,-Rsinα,z).  For r-r1  × dr1  we have (zcosα,zsinα,R)Rdα.

|rr1|3=(R2+z2)3/2

For

B=μ0I4π R(R2+z2)3/202π(osα,zsinα,R)dα=μ0I4π R(R2+z2)3/2 2πRez=μ0IR2(R2+z2)3/2 ez
Figure a. Magnetic field for a circled wire
Figure a. Magnetic field for a circled wire
Example 2. Determine the magnetic field B(r) of solenoid
A solenoid is a conducting coil, and it’s length is usually several times bigger than the radius. The structure of a magnetic field of a solenoid is the following: as the coil consist of a large quantity of ‘current turns’, each one makes an impact into the resulting magnetic field. And for every turn, there is a symmetrical turn according to the plane, perpendicular to the coil axis. The sum of inductions B(r) for symmetrical coils are parallel to the coil axis, and the magnetic field occurs parallel to the coil axis inside and outside of the coil. Let’s choose the contour abcd, as it is shown in the Figure b. Circulation of vector B for this contour is

B=abBldl+bcBldl+cdBldl+daBldl=Bl

then

Bl=μ0iIi=μ0NI,B=μ0nI

Magnetic fields outside of the solenoid are very small, and nearly all the magnetic fields ars concentrated inside of it. The magnetic field of the solenoid is uniform.

Figure b. Magnet field in the solenoid
Figure b. Magnet field in the solenoid

Sometimes wires form complex structures. So how do we use the Biot-Savart-Laplace Law in this case? To resolve these cases the structure can be divided into segments. And Biot-Savart-Laplace Law can be applied individually to each segment. Then for a complex structure we get:

B=μ0I4πirdrr3

Let’s consider the segment of a wire depicted on the Figure c. rdr=dhdl, where h is the line from the point where we are estimating the magnetic field to the wire segment, or its extrapolation. l is coinciding with the wire segment. As h is perpendicular to the wire segment, then hxl=h*l. And drxr=dhxdl. Then

B=μ0I4πl1l2 hdlr3

Here

r2=l2+h2,and r3=(l2+h2)3/2

So

B=μ0I4πl1l2 hdl(l2+h2)3/2

Let’s express l1 and l2 with known variables – h and α.

l=hcotα,dl=hdαsin2α),l2+h2=h2sin2α),hdl(l2+h2)3/2=hhdαsin2α sin3αh3=dαsinαh

Then

l1l2 hdl(l2+h2)3/2=1hα1α2 dcosα=cosα1cosα2h

For the given straight segment of wire

B=μ0I4πcosα1cosα2hez
Figure c. Magnet field calculation for a wire fragment
Figure c. Magnet field calculation for a wire fragment
Example 3. Let’s calculate the magnetic field for the circuit in Figure d

The square sides are d, the

α1=45,α2=135

Magnetic field at the centre of the square

B=μ0I2πd(cos45cos135)=22π μ0Id

(magnetic field for the triangle).

B=22π μ0Id
Figure d. The wire segments, shaped as a square
Figure d. The wire segments, shaped as a square

Electromagnetic induction

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