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 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.

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 = … … .

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

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

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.

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

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

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.