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Magnetic field at an axial point due to current in circular wire

Module by: Sunil Kumar Singh. E-mail the author

We have already determined magnetic field due to current in circular wire at its center. The approach to determine magnetic field at an axial point is similar. We begin with magnetic field due to small current element and then try to integrate the Biot-Savart expression for the small magnetic field for the entire circle following superposition principle.

This extension of earlier procedure, however, demands a bit of extra three dimensional imagination to arrive at the correct result. In this module, we shall attempt to grasp three dimensional elements as clearly as possible with figures. Let us have a look at the differential Biot-Savart expression :

đ B = μ 0 4 π I đ l X r r 3 đ B = μ 0 4 π I đ l X r r 3

There are three vector quantities dB, dl and r. We investigate the spatial relation among these quantities for magnetic field at an axial point.

Magnetic field on an axial point

The magnitude of magnetic field due to current in a current element is given by :

đ B = μ 0 4 π I đ l sin θ r 2 đ B = μ 0 4 π I đ l sin θ r 2

In order to evaluate magnetic field due to complete circular wire, we need to set up corresponding integral properly with respect to various elements constituting the expression. In following subsections, we study these elements in which point of observation is a point on axial line.

The angle between current length element and displacement vectors

The angle (θ ) as appearing in the Biot-Savart expression between current length element vector dl and displacement vector r is right angle. See figure. This right angle should be distinguished with acute angle φ, which is the angle between OA and AP as shown in the figure.

Figure 1: The angle between current length element and displacement vectors is right angle.
The angle between current length element and displacement vectors
The angle between current length element and displacement vectors (cl1.gif)

The above fact reduces Biot-Savart expression to :

đ B = μ 0 4 π I đ l sin 90 r 2 = μ 0 4 π I đ l r 2 đ B = μ 0 4 π I đ l sin 90 r 2 = μ 0 4 π I đ l r 2

This simplification due to enclosed angle being right angle is true for all points on the circle.

Magnitude of magnetic field

All current elements are at equal linear distance from point P. As a result, the magnitude of magnetic field at P due to any of the equal current elements is same.

đ B 1 = đ B 2 = … … . đ B 1 = đ B 2 = … … .

Direction of elemental magnetic field

Unlike enclosed angle (θ), linear distance (r) and magnitude of magnetic field, the direction of magnetic field due to current elements are not same. As such, we can not integrate Biot-Savart differential expression to determine net magnetic field at P. Let us investigate the direction of magnetic fields due to two diametrically opposite current elements. Let the circular wire lie in yz plane as shown in the figure.

Figure 2: Magnetic field is perpendicular to plane formed by current length element and displacement vectors.
Direction of elemental magnetic field
Direction of elemental magnetic field (cl2.gif)

The current length vector d l 1 d l 1 and displacement vector r 1 r 1 form a plane shown as plane 1 and the magnetic field due to current element, B 1 B 1 , is perpendicular to plane 1. Similarly, the current length vector d l 2 d l 2 and displacement vector r 2 r 2 form a plane shown as plane 2 and the magnetic field due to current element, B 2 B 2 ), is perpendicular to plane 2. Clearly, these magnetic fields are directed in three dimensional space. If we imagine magnetic fields due to other current elements of the circular wire, then it is not difficult to imagine that these elemental magnetic fields are aligned on the outer surface of a conic section and that they are not in same plane.

Figure 3: Magnetic fields are aligned on the outer surface of a conic section.
Direction of elemental magnetic field
Direction of elemental magnetic field (cl5.gif)

Another important point to observe is that all elemental magnetic field vectors form same angle φ. This can be verified from the fact that B 1 B 1 is perpendicular to AP and Px is perpendicular to OA. Hence, angle between B 1 B 1 and Px is equal to angle between OA and AP i.e. φ with x-axis. By symmetry, we can see that all elemental magnetic field vectors form the same angle with x- axis.

Resolution of elemental magnetic field vectors and net magnetic field

We resolve magnetic field vectors along x-axis and perpendicular to it, which lies on a plane perpendicular to axis i.e a plane parallel to the plane of circular coil (yz plane) as shown in the figure. We have shown two pairs of diametrically opposite current elements. See that axial components are in positive x-direction. The perpendicular components, however, cancels each other for a diametrically opposite pair.

Figure 4: Net magnetic field is axial.
Resolution of elemental magnetic field vectors and net magnetic field
Resolution of elemental magnetic field vectors and net magnetic field (cl3.gif)

This situation greatly simplifies the integration process. We need only to algebraically add axial components. Since all are in same direction, we integrate the axial component of differential Biot-Savart expression :

B = ∫ đ B = μ 0 I 4 π ∫ đ l r 2 cos φ B = ∫ đ B = μ 0 I 4 π ∫ đ l r 2 cos φ

Note that both r and cos φ are constants and they can be taken out of integral,

⇒ B = μ 0 I cos φ 4 π r 2 ∫ đ l ⇒ B = μ 0 I cos φ 4 π r 2 ∫ đ l ⇒ B = μ 0 I cos φ 4 π r 2 X 2 π R = μ 0 I R cos φ 2 r 2 ⇒ B = μ 0 I cos φ 4 π r 2 X 2 π R = μ 0 I R cos φ 2 r 2

Now,

r = x 2 + R 2 1 2 r = x 2 + R 2 1 2

In triangle OAP,

cos φ = R r = R x 2 + R 2 1 2 cos φ = R r = R x 2 + R 2 1 2

Putting these values in the expression of magnetic field, we have :

⇒ B = μ 0 I R cos φ 2 r 2 = μ 0 I R 2 2 x 2 + R 2 3 2 ⇒ B = μ 0 I R cos φ 2 r 2 = μ 0 I R 2 2 x 2 + R 2 3 2

This is the expression of magnitude of magnetic field on axial line. Note that we have derived this expression for anticlockwise current. For clockwise current, the magnetic field will have same magnitude but oriented towards the circular wire. Clearly, direction of axial magnetic field follows Right hand thumb rule.

If there are N turns of circular wires stacked, then magnetic field is reinforced N times and magnetic field is :

⇒ B = μ 0 N I R 2 2 x 2 + R 2 3 2 ⇒ B = μ 0 N I R 2 2 x 2 + R 2 3 2

In order to show the direction, we may write the expression for magnetic field vector using unit vector in the axial direction as :

⇒ B = μ 0 N I R 2 2 x 2 + R 2 3 2 i ⇒ B = μ 0 N I R 2 2 x 2 + R 2 3 2 i

Recall that one of the faces of circular wire has clockwise direction of current, whereas other face of the same circular wire has anticlockwise direction of current. The magnetic field lines enter from the face where current is clockwise and exit from the face where current is anticlockwise.

Example 1

Problem : Two identical circular coils of radius R are placed face to face with their centers on a straight line at a distance 2√ 3 R apart. If the current in each coil is I flowing in same direction, then determine the magnetic field at a point “O” midway between them on the straight line.

Figure 5: Two identical circular coils at a distance
Two identical circular coils at a distance
Two identical circular coils at a distance (cl6a.gif)

Solution : For an observer at “O”, the current in coil A is anticlockwise. The magnetic field due to this coil is towards the observer i.e. towards right. On the other hand, the current in coil C is clockwise for an observer at “O”. The magnetic field due to this coil is away from the observer i.e. again towards right. The magnitude of magnetic field due to either coil is :

B ′ = μ 0 I R 2 2 x 2 + R 2 3 / 2 B ′ = μ 0 I R 2 2 x 2 + R 2 3 / 2

Here, x = √3 R,

⇒ B ′ = μ 0 I R 2 2 3 R 2 + R 2 3 / 2 = μ 0 I R 2 2 4 R 2 3 / 2 = μ 0 I 16 R ⇒ B ′ = μ 0 I R 2 2 3 R 2 + R 2 3 / 2 = μ 0 I R 2 2 4 R 2 3 / 2 = μ 0 I 16 R

The net magnetic field is twice the magnetic field due to one coil,

⇒ B = 2 B ′ ⇒ B = 2 B ′ ⇒ B = 2 B ′ = 2 X μ 0 I 16 R = μ 0 I 8 R ⇒ B = 2 B ′ = 2 X μ 0 I 16 R = μ 0 I 8 R

The net magnetic field is directed towards right.

Variation of magnetic field along the axis

As far as magnitude of magnetic field is concerned, it decreases away from the circular wire. It is maximum when point of observation is center. In this case,

x = 0 and magnetic field, B is :

B = μ 0 I R 2 2 x 2 + R 2 3 2 = μ 0 I R 2 2 R 3 = μ 0 I 2 R B = μ 0 I R 2 2 x 2 + R 2 3 2 = μ 0 I R 2 2 R 3 = μ 0 I 2 R

This result is consistent with the one derived for this case in earlier module. For magnetic field at a far off point on the axis,

x 2 ≫ R 2 x 2 ≫ R 2 x 2 + R 2 ≈ x 2 x 2 + R 2 ≈ x 2

Putting in the expression of magnetic field, we have :

B = μ 0 I R 2 2 x 2 + R 2 3 2 = μ 0 I R 2 2 x 3 B = μ 0 I R 2 2 x 2 + R 2 3 2 = μ 0 I R 2 2 x 3

Clearly, magnetic field falls off rapidly i.e. inversely with the cube of linear distance x along the axis. A plot of the magnitude of current is shown here in the figure :

Figure 6: Variation of magnetic field along the axis
Variation of magnetic field along the axis
Variation of magnetic field along the axis (cl4.gif)

Magnetic moment

The concept of moment is a very helpful concept for describing magnetic properties. The description of circular coil as magnetic source in terms of magnetic moment, as a matter of fact, underlines yet another parallelism that runs between electrostatics and electromagnetism.

Magnetic moment of a closed shaped wire is given by :

M = N I A M = N I A

For a single turn of circular wire :

M = I A M = I A

The magnetic moment is a vector obtained by multiplying area vector with current. The direction of area vector is perpendicular to the plane of wire. For circular wire shown in the module,

A = π R 2 i A = π R 2 i

Now, axial magnetic field vector is given by :

B = μ 0 I R 2 2 x 2 + R 2 3 2 i B = μ 0 I R 2 2 x 2 + R 2 3 2 i

But,

M = I A = I π R 2 i M = I A = I π R 2 i

Substituting in the expression of magnetic field,

⇒ B = μ 0 M 2 π x 2 + R 2 3 2 ⇒ B = μ 0 M 2 π x 2 + R 2 3 2

For a far off axial point ( x 2 + R 2 ≈ x 2 x 2 + R 2 ≈ x 2 ):

⇒ B = μ 0 M 2 π x 3 = μ 0 2 M 4 π x 3 ⇒ B = μ 0 M 2 π x 3 = μ 0 2 M 4 π x 3

See the resemblance; it has the same form as that for electrical field due to an electrical dipole having dipole moment pon an axial point :

E = 2 p 4 π ε 0 x 3 E = 2 p 4 π ε 0 x 3

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