0.5 Magnetic field at an axial point due to current in circular  (Page 2/2)

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 $B_1$ is perpendicular to AP and Px is perpendicular to OA. Hence, angle between $B_1$ 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.

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 :

$$B=\int đB=\frac{{\mu }_{0}I}{4\pi }\int \frac{đl}{r^2}\cos \phi$$

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

$$\Rightarrow B=\frac{{\mu }_{0}I\cos \phi }{4\pi r^2}\int đl$$ $$\Rightarrow B=\frac{{\mu }_{0}I\cos \phi }{4\pi r^2}X2\pi R=\frac{{\mu }_{0}IR\cos \phi }{2r^2}$$

Now,

$$r={\left(x^2+R^2\right)}^{\frac{1}{2}}$$

In triangle OAP,

$$\cos \phi =\frac{R}{r}=\frac{R}{{\left(x^2+R^2\right)}^{\frac{1}{2}}}$$

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

$$\Rightarrow B=\frac{{\mu }_{0}IR\cos \phi }{2r^2}=\frac{{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}$$

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 :

$$\Rightarrow B=\frac{{\mu }_{0}NI{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}$$

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

$$\Rightarrow \mathbf{B}=\frac{{\mu }_{0}NI{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}\mathbf{i}$$

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.

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.

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 :

$$B\prime =\frac{{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{3/2}}$$

Here, x = √3 R,

$$\Rightarrow B\prime =\frac{{\mu }_{0}I{R}^2}{2{\left(3R^2+R^2\right)}^{3/2}}=\frac{{\mu }_{0}I{R}^2}{2{\left(4R^2\right)}^{3/2}}=\frac{{\mu }_{0}I}{16R}$$

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

$$\Rightarrow B=2B\prime$$ $$\Rightarrow B=2B\prime =2X\frac{{\mu }_{0}I}{16R}=\frac{{\mu }_{0}I}{8R}$$

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 :

$$B=\frac{{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}=\frac{{\mu }_{0}I{R}^2}{2R^3}=\frac{{\mu }_{0}I}{2R}$$

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,

$$x^2\gg R^2$$ $$x^2+R^2\approx x^2$$

Putting in the expression of magnetic field, we have :

$$B=\frac{{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}=\frac{{\mu }_{0}I{R}^2}{2x^3}$$

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 :

Variation of magnetic field along the axis

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 :

$$\mathbf{M}=NI\mathbf{A}$$

For a single turn of circular wire :

$$\mathbf{M}=I\mathbf{A}$$

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,

$$\mathbf{A}=\pi R^2\mathbf{i}$$

Now, axial magnetic field vector is given by :

$$\mathbf{B}=\frac{{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}\mathbf{i}$$

But,

$$\mathbf{M}=I\mathbf{A}=I\pi R^2\mathbf{i}$$

Substituting in the expression of magnetic field,

$$\Rightarrow \mathbf{B}=\frac{{\mu }_0\mathbf{M}}{2\pi {\left(x^2+R^2\right)}^{\frac{3}{2}}}$$

For a far off axial point ( $x^2+R^2\approx x^2$ ):

$$\Rightarrow \mathbf{B}=\frac{{\mu }_0\mathbf{M}}{2\pi x^3}=\frac{{\mu }_{0}2\mathbf{M}}{4\pi x^3}$$

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

$$\mathbf{E}=\frac{2\mathbf{p}}{4\pi {\epsilon }_0{x}^3}$$