## Magnetic Field Strength Outside a Long Current Carrying Conductors:

Consider a straight long conductor carrying current of I amperes in downward direction. Let the conductor be influenced only by the field produced by the current flowing through it (no external filed). Let the field strength at any point at a distance of r meters from the centre of the conductor due to its own filed be H newton/wb.

For acting on a unit N-pole placed at this point = H newtons, tangential to the lines of force

Work done in moving a unit N-pole along the circumstance of a circle of radius r meters against this force = Force distance moved

joules …..(i)

But a unit N-pole emanates a flux one weber, therefore, in one round of conductor, flux of one weber is cut by the conductor, hence work done is given as

Work done = Flux cut by the conductor current flowing through the conductor = …..(ii)

Equation (i) and (ii) we get

Or ampere-turns/meter ……(1)

And flux density,

tesla …..(2)

Obviously, if there are N-conductor as shown in figure 2, then

AT/m …..(3)

tesla….in air …..(4)

tesla in a substance of relative permeability …..(5)

To determine the magnetizing force inside the conductor, draw a circular section of radius r and assume that the current density is same at any point of this x-section and equal to where a is the radius of conductor. Then the total current threading through the encircled cross-section area can be found as

Hence it may be written that

When from …..(6)

Or

The curve of the figure 3 plotted in accordance with equations (6) and (1) shows the variation of conductor with the distance from the axis of the conductor.

### Field Strength of a Closely Wound Toroidal Coil:

The path in this is a circle with a radius r. the path links N turns carrying current of the same direction.

According to Ampere’s work law

Or …..(7)

Or

Therefore, l, length of toroidal core =

Thus, the strength of magnetic field due to a toroid is directly proportional to the ampere-turns, or magneto-motive force (mmf) F, and inversely proportional to the length of the path of the magnetic flux.

The magnetic induction (or magnetic flux density) inside a toroid will be

tesla(or Wb/m^{2}) in a medium …..(8)

tesla(or Wb/m^{2}) in air …..(9)

## Magnetic Field Strength Due to Finite Length of Wire Carrying Current

Consider a straight wire of length l carrying a steady current I. Let us find magnetic field strength H at a point P at a distance R from the wire, as shown in figure 5

The magnetic induction due to small element dl of the wire shown in figure 2 is

By Biot-Savart law

Or

Therefore and are complementary angles.

Magnetic field strength due to entire length.

From figure 2

or

Therefore R is constant and from figure 5

Or

Now from figure 5

Therefore,

And

And so limits get transformed from 0 to .

So,

…..(10)

For wire of infinity length extended it at both ends (i.e. the limit of integration would be giving

……(11)

## Magnetic Field Strength Along the Axis of a Square Coil

This is a similar case as discussed above except that there are four conductors each of length, say, 2a meters, and carrying a current of I amperes as shown in Figure 6. The magnetic field strengths at the axial-point P due to opposite sides AB and CD, H_{AB} and H_{CD}, are directed at right angles to the planes containing P and AB and P and CD respectively.

Since H_{AB} and H_{CD} are numerically equal, therefore, their components at right angles to the axis of the coil will cancel out but the axial components will add. Similarly, the other two sides will also give a resultant axial component only.

The magnetic field strength due to side AB is

Or

Therefore, from figure 6

Axial components of

All the four sides pf the square coil will contributes an equal amount to the resulting magnetic field at point P. So, magnetic field strength at point P,

Substituting and in above expression we get

AT/m …..(12)

At the centre of the square coil, x = 0 and

AT/m …..(13)

The field strength at the centre of a square coil can also be determined directly as under. As seen from figure 7, the magnetic field intensity at point O due to any side is given as

Or

Resulting magnetic field strength due to all four sides is

AT/m, the same in equation (13)

## Magnetic Field Strength on the Axis of a Circular Coil

Consider a single turn circular coil of radius a meters and carrying a current of I amperes.

The magnetic field strength at any point P lying on the axis of a circular coil due to a small element dl at point A, as given by Biot Savart law, is

newton along PQ where r is the distance of point P from small length dl. Now dH can be resolved into two components PA and PB respectively along and perpendicular to the coil axis.

Similar magnetizing force at point P due to diametrically opposite small element dl’ at point B can also be resolved into two components PA and PC along axis PO and perpendicular to axis PO respectively. The vertical components due to small elements dl at points A and B are equal in magnitude but opposite in direction and so cancel each other.

Thus on considering the whole of the coil, the sum of vertical components will be zero and the magnetizing force at point P will be equal to sum of all the axial components.

So, magnetizing force at point P due to whole of the coil

Therefore,

Or AT/m …..(14)

Therefore, and

At the centre of the coil, x = 0

Therefore, Field strength, H = \dfrac{I}{2a}[/latex] AT/m …..(15)

If there are N turns on the coil then the equation (14) and (15) become

AT/m …..(16)

And AT/m …..(17)

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