Raghavendra G. Kulkarni
The classical method of obtaining an expression for the magnetic field due to an infinite current sheet employs Ampere’s circuital law. The method is well covered in introductory textbooks on electromagnetic field theory; see [1], [2], [3], and [4]. The magnetic field (H) at any point due to an infinite current sheet with a surface current density of K A/m is given by \[{\bf{H}} = \frac{1}{2}{\bf{K}}\,{\times}\,{\bf{a}}_{N} \tag{1} \] where ${\bf{a}}_{N}$ is the unit normal of the current sheet pointing toward the field point.
In this article, we present an alternative method to derive the result shown in (1) without employing Ampere’s circuital law, but using instead the Biot–Savart law, which is generally not discussed in textbooks since the derivation of the magnetic field using this method is more elaborate than the method using Ampere’s circuital law. However, the method using Ampere’s circuital law can be employed only when we have some prior information about the magnetic field; for example, it has symmetry, or it is constant along the closed path chosen. The method using the Biot–Savart law is the more general one, and it can be employed in all cases. Therefore, it is expected that the method (using the Biot–Savart law) described here will provide more insight to the reader.
Consider an infinite current sheet in the xy-plane, with current flowing in the y direction. The sheet has a surface current density of ${\bf{K}} = {K}_{y}{\bf{a}}_{y}$ A/m (see Figure 1).
Figure 1. Infinite current sheet in xy-plane with ${\bf{K}} = {K}_{y}{\bf{a}}_{y}$.
Let us imagine that the current sheet is made up of an infinite number of infinitely long filaments (of infinitesimal width dx), placed side by side along the x direction, each carrying a current of ${K}_{y}{dx}$. Now, consider a field point P on the z-axis at a distance h from the current sheet in the region ${z}\,{>}\,{0}$, and a filament on the +ve x-axis at a distance x from the origin (see Figure 2). So, the distance of the field point P from the filament is ${\left\vert{R}_{1}\right\vert} = {R}_{1} = \sqrt{{x}^{2} + {h}^{2}}$. The magnetic field at P due to the filament is \[{d}{\bf{H}}_{1} = \frac{{K}_{y}dx}{{2}{\pi}{R}_{1}}{\bf{a}}_{1} \tag{2} \]
Figure 2. Infinite current filament on +ve x-axis carrying a current of ${K}_{y}{dx}$.
where ${\bf{a}}_{1}$ is the unit vector at P in the direction of the curl, ${\bf{K}}\,{\times}\,{\bf{R}}_{1}$. From Figure 2, notice that ${\bf{a}}_{1} = {\sin}\,{\alpha}{\bf{a}}_{x} + {\cos}\,{\alpha}{\bf{a}}_{z}$. Hence (2) becomes \[{d}{\bf{H}}_{1} = \frac{{K}_{y}dx}{{2}{\pi}{R}_{1}}{(}{\sin}\,{\alpha}{\bf{a}}_{x} + {\cos}\,{\alpha}{\bf{a}}_{z}{)}. \tag{3} \]
Now, consider the second filament on the negative x-axis at a distance x from the origin (Figure 3); it is the mirror of the first filament. The distance of point P from the second filament is ${R}_{2} = \sqrt{{x}^{2} + {h}^{2}}$. Note that ${R}_{2} = {R}_{1}$, but the vectors ${\bf{R}}_{1}$ and ${\bf{R}}_{2}$ are not the same. The magnetic field at P due to the second filament is \[{d}{\bf{H}}_{2} = \frac{{K}_{y}dx}{{2}{\pi}{R}_{2}}{\bf{a}}_{2} = \frac{{K}_{y}dx}{{2}{\pi}{R}_{1}}{\bf{a}}_{2} \tag{4} \]
Figure 3. Infinite current filament on –ve x-axis carrying a current of ${K}_{y}{dx}$.
where ${\bf{a}}_{2}$ is the unit vector at P in the direction of the curl, ${\bf{K}}\,{\times}\,{\bf{R}}_{2}$. From Figure 3, notice that ${\bf{a}}_{2} = {\sin}\,{\alpha}{\bf{a}}_{x}{-}{\cos}\,{\alpha}{\bf{a}}_{z}$. Hence, (4) becomes \[{d}{\bf{H}}_{2} = \frac{{K}_{y}dx}{{2}{\pi}{R}_{1}}{(}{\sin}\,{\alpha}{\bf{a}}_{x}{-}{\cos}\,{\alpha}{\bf{a}}_{z}{)}. \tag{5} \]
The total magnetic field due to the first filament and the second filament is the vector sum of the individual magnetic fields, ${d}{\bf{H}} = {d}{\bf{H}}_{1} + {d}{\bf{H}}_{2}$. Note that the z components of the magnetic fields cancel and the x components add up, yielding \[{d}{\bf{H}} = \frac{{K}_{y}dx}{{\pi}{R}_{1}}{\sin}\,{\alpha}{\bf{a}}_{x}{.} \tag{6} \]
From Figure 2 we note that ${\sin}\,{\alpha} = {h} / {R}_{1}$; using this in (6) results in \[{d}{\bf{H}} = \frac{{K}_{y}hdx}{{\pi}{R}_{1}^{2}}{\bf{a}}_{x} = \frac{{K}_{y}hdx}{{\pi}{(}{x}^{2} + {h}^{2}{)}}{\bf{a}}_{x}{.} \tag{7} \]
The total magnetic field due to the entire current sheet is obtained by integrating expression (7) from ${x} = {0}$ to ${x} = {\infty}$: \[{\bf{H}} = {\bf{a}}_{x} \mathop{\int}\nolimits_{0}\nolimits^{\infty} \frac{{K}_{y}hdx}{{\pi}{(}{x}^{2} + {h}^{2}{)}}{.} \tag{8} \]
Evaluation of the integral in (8) yields an expression for the magnetic field in the region ${z}\,{>}\,{0}$: \[{\bf{H}} = {\bf{a}}_{x} \frac{1}{2}{K}_{y}{.} \tag{9} \]
Notice from (9) that the magnetic field is independent of the distance from the current sheet. Similarly, in the region ${z}\,{<}\,{0}$, the magnetic field is \[{\bf{H}} = {-}{\bf{a}}_{x} \frac{1}{2}{K}_{y}{.} \tag{10} \]
Generalizing expressions (9) and (10) for the magnetic field in any region, we get \[{\bf{H}} = \frac{1}{2} {\bf{K}}\,{\times}\,{\bf{a}}_{N} \tag{11} \] which is the same as (1).
In this article, we have obtained the magnetic field due to an infinite current sheet using the Biot–Savart law, which is an alternative method to the classical method employing Ampere’s circuital law normally described in textbooks. While the use of Ampere’s circuital law yields an easier derivation of the magnetic field, it requires that the unknown magnetic field must be taken out of the closed line integral, implying that the magnetic field must be constant along the path chosen. Thus, this method is applicable only for certain symmetric cases. In the case of the Biot–Savart law, the integration is performed on a function comprising known quantities, such as an infinitesimal current element and the distance between the field point and current element. Therefore, this method is a more general one, and it can be employed in all cases; but the derivation becomes lengthier than that using Ampere’s circuital law. The author feels that an awareness of the two methods will be a rewarding experience to the students of electromagnetics.
The author thanks the editorial board of IEEE Microwave Magazine for providing tips to improve the original manuscript. This work is supported by PES University, Bengaluru.
[1] E. C. Jordan and K. G. Balmain, Electromagnetic Waves and Radiating Systems, 2nd ed. New Delhi, India: Prentice-Hall, 2009.
[2] W. H. Hayt Jr. and J. A. Buck, Engineering Electromagnetics, 8th ed. New York, NY, USA: McGraw-Hill, 2012.
[3] M. N. O. Sadiku, Elements of Electromagnetics, 6th ed. New York, NY, USA: Oxford Univ. Press, 2015.
[4] S. W. Ellingson, Electromagnetics, vol. 1. Blacksburg, VA, USA: VT Publishing, 2018.
Digital Object Identifier 10.1109/MMM.2023.3284773