Sources of the magnetic field

Oersted discovered in 1819 that a compass needle was deflected near a conducting wire carrying a current. The electric current was the source of the magnetic field that exerted a torque on the compass needle. This observation was the first indication of the connection between electricity and magnetism.

In this section, we will give the Biot-Savart law and Ampere's law and some applications of them.

1. The biot-Savart law

The discovery of Oersted was followed by the experiments by Ampère and by Biot and Savart that led to their laws.

In a point in space, the Boit-Savart law determines the magnetic field due to a distribution of electric currents.

Let's consider a current distribution where the current element Idl (source of dB) produces the infinitesimal contribution dB to the magnetic field at the point P distant of r from the origin of the vector current element dl (point source).

The direction of the vector dB is given by the direction of the cross product Idl x r. It is directed out of the page toward us ⊙. It is perpendicular to both dl and ^r, that is to the plane formed by Idl and ^r. The sense of the vector db is given by the right-hand rule. ^r is the unit vector pointing from the current element to the point P. The Bio-Savart law for an infinitesimal current element is:

Integrated along the line of the current distribution, Biot-Savart law in its integral form is:

The magnetic field at a point is the linear superposition 0f the vector contributions due to each of the infinitesimal current elements.

The constant μ_o is called the permeability constant for vacuum. Its value in the SI units is given by:

μ_o/4π = 1.00 x 10^{- 7} T.m/A

2. Magnetic field and Electric field analogy

For electromagnetism, the Biot-Savart law is analogous to Coulomb's law for electrostatics. Coulomb's law gives the electric field produced by a distribution of charges. If dq is a differential of infinitesimal element of charge, then the electric field dE produced at a point P by that charge is given by the equation:

r is the distance from the element of charge to the point P and ^r is the unit vector directed from the charge to the point P. Integrating over the charge distribution gives the electric field at the point P:

Now, a current distribution with a current element Idl produces a contribution dB to the magnetic field at a point P. r is the distance from the current element(source) to point P. The Boit-Savart law for a differential current element is:

Integrated along the line of the current distribution, Biot-Savart law in its integral form is:

The direction of E is radial with respect to the charge dq while the direction of B is perpendicular to the plane formed by the vector dl and the the vector r.

3. Magnetic field due to a current
in a long straight wire

We want to determine the magnetic field produced by a long straight wire crossed by a current I, where the point P is in the x-axis and at a perpendicular distance from the wire. The direction of dB is given by the right-hand rule, so it is directed into the page. The magnitude of dB is:

dB = (μ_o/4Ï€) I dx sin θ /r²

sin θ = sin (Ï€ - θ) = R/r = R/[x² + R²]^1/2

dB = (μ_o/4Ï€) I dx R /r³ =
(μ_oIR/4Ï€) dx /[x² + R²]^3/2

Therefore

B = (μ_oIR/4Ï€) ∫ dx /[x² + R²]^3/2
-∞ → + ∞

= (2μ_oIR/4Ï€) ∫ dx /[x² + R²]^3/2
0 → + ∞

I = ∫ dx /[x² + R²]^3/2
Let x = R tan z. So dx = R (1 + tg²z) dz =
R (1 + (x/R²) dz

I = ∫ dz R (1 + (x/R²) /[(x/R)² + 1]^3/2R³ =
(1/R²) ∫ dz/[tan²z + 1]^1/2 = (1/R²) ∫ cos z dz =
sin z + const.

sin z = sin (Arctg(x/R))
Arctg(x/R) = α → x/R = tan α

We have α = θ - π/2. Hence
sin z = sin (θ - π/2) = - cos θ = cos(π - θ) =
x/r = x/[x² + R²]^1/2

Therefore

I = (1/R²) . x/[x² + R²]^1/2 = (1/R²)(1 - 0) = (1/R²)
0 → + ∞

B = (2μ_oIR/4Ï€) . (1/R²)

B = (μ_oI/2πR)

4. Magnetic field due to a current loop

Consider the magnetic field produced by a circular loop of radius R carrying a current I. We want to determine the magnetic field on a point P along the axis of the loop, x-axis. Any point on the loop is equidistant to that point P. That is for a given point P on the x-axis, then a given distance x, r remains constant. The contribution dB to the field at this point is:

dB = (μ_o/4Ï€) Idl/r² that we resolve into the components parallel and perpendicular to the x-axis. From symmetry, the pependicular components add to zero. The parallel components give :

dB_∥ = (μ_o/4Ï€) Idl cos θ/r² = (μ_o/4Ï€) Idl/r² . (R/r) =
(μ_o/4Ï€) IRdl/r³

Therefore

B = ∫dB dl = (μ_o/4Ï€) IR/r³ ∫ dl = (μ_o/4Ï€) IR/r³ (2Ï€R) =

(μ_o/2) IR²/r_{3 = (μoIR²)/2r³}