In turn, the circulation of B is the sum of all products between the tangential component B ║ and the length of a small segment Δℓ of a closed curve C , around a circuit. In mathematical terms, it is written like this:
∑ B ║ .Δℓ ∝ I
Like an arbitrary line or curve C, it can be divided into small segments Δℓ , which in turn can be infinitesimal, so they are called d ℓ .
In this case, the sum becomes a line integral of the dot product between the vectors B and d s. This product contains the tangential component of B, which is B cosθ, where θ is the angle between the vectors:
The small circle that the integral travels means that the integration takes place in a closed path C, which in this case involves current flowing through the conductor’s cross section.
The proportionality constant necessary to establish equality is µ o , the vacuum permeability. In this way, Ampère’s law remains:
Ampère’s law tells us that the line integral ∫ C B ∙ d s is exactly μ or I, but it does not give us details on how the magnetic field B is oriented with respect to the curve C at each point, nor on how to calculate the full. It just tells us that the result is always μ or I.
Demonstration of Ampère’s Law
Ampère’s law is experimentally verified by checking the magnetic field produced by a very long straight conductor. Before approaching the problem, we must highlight two cases of special interest in the previous equation:
-The first one is when B d s are parallel, which means that B is tangential to C. Then the angle between the two vectors is 0 and the dot product is simply the product of the quantities B.ds .
-The second occurs if B and d s are perpendicular, in which case the dot product is 0, since the angle between the vectors is 90º, whose cosine is 0.
Another important detail is the choice of curve C in which the field circulation is evaluated. Ampère’s law does not specify what it can be, but it must involve the actual distribution. It also does not say which way to take the curve and there are two possibilities for that.
The solution is to assign signals according to the right thumb rule. The four fingers are curved in the direction you want to integrate, usually the same as the B- field circles. If the current points in the direction of the right thumb, it is assigned a + sign, and if not, a – sign.
This applies when there is a distribution with several currents, some may be positive and some negative. Their algebraic sum is what we’re going to put in Ampère’s law, which is usually called a closed current (by the curve C).
Infinite Straight Wire Magnetic Field
Figure 2 shows a wire that carries a current I out of the plane. The right thumb rule ensures that B circles counterclockwise, describing the circles as shown by the red arrows.
Let’s take one of them, whose radius is r. We divide into small differential segments d s , represented by the vectors in blue. Both vectors, B and d s , are parallel at each point of the circle and therefore the integral ∫ C B ∙ d s becomes:
∫ C Bds
This is because, as we said before, the scalar product B ∙ d s is the product of the magnitudes of the vectors by the cosine of 0º. We know the result of the integral thanks to Ampère’s law, so we write:
∫ C Bds = μ or I
As the magnitude of the field is constant all the way through, it leaves the integral:
B ∫ C ds = μ or I
The integral ∫ C ds represents the sum of all infinitesimal segments that make up the circumference of radius r , equivalent to its length, the product of its radius in 2π:
B.2πr = μ or I
And from there, we find that the magnitude of B is:
B = μ or I / 2πr
It should be emphasized that even if the selected path ( or ampere circuit ) were not circular, the result of the integral is still μ or I, however, ∫ C B ∙ d would no longer be B.2πr.
Therefore, the usefulness of Ampère’s law to determine the magnetic field lies in choosing distributions with high symmetry, so that the integral is easy to evaluate. Circular and straight paths meet this requirement.
– Exercise 1
Consider the curves a, b, c and shown in Figure 3. They involve three currents, two leaving the plane, symbolized by a dot ( . ), whose intensities are 1 A and 5 A and a current entering the plane, which is indicated by a cross and whose magnitude is 2 A.
Find the current involved by each curve.
Chains exiting the paper receive a + sign. According to this:
It includes all three currents, so the included current is +1 A + 5 A – 2 A = 4 A.
Only 1 A and – 2 A currents are within this curve; therefore, the included current is – 2 A.
It includes 1A and 5A output currents, so the included current is 6A.
The internal currents are +5 A and – 2 A, so it contains a net current of 3 A.
– Exercise 2
Calculate the magnitude of the magnetic field produced by a very long straight wire, at a point of 1 meter, if the wire carries a current of 1 A.
According to Ampère’s law, the field of the wire is given by:
B = μ or I / 2πr = (4π x 10 -7 x 1 / 2π x 1) T = 2 x 10 -7 T.