# Ampère’s Law: Formula And Equations, Proof, Exercises

The **Ampere ‘s law** states that the movement of the magnetic induction vector **B** is proportional to the intensity I of the current flowing through the same.

In turn, the circulation of **B** is the sum of all the 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*

As an arbitrary line or curve C, it can be divided into small segments *Δℓ* , and these in turn can be infinitesimal, then they are called d *ℓ* .

In such a case, the summation becomes a line integral of the scalar product between vectors **B** and d **s. **Said product contains the tangential component of B, which is B cosθ, where θ is the angle between the vectors:

The small circle through the integral means that the integration takes place over a closed path C, which in this case involves the current flowing through the cross section of the conductor.

The constant of proportionality necessary to establish equality is μ _{or} , the permeability of the vacuum. In this way, Ampère’s law is:

Ampère’s law tells us that the line integral ∫ _{C } **B** ∙ d **s is** worth exactly μ _{or} I, but it does not give us the details about how the magnetic field **B** is oriented with respect to the curve C at each point, nor about the way to calculate the integral. It only tells us that the result of it is always μ _{or} I.

__Demonstration of Ampère’s law__

__Demonstration of Ampère’s law__

Ampère’s law is verified experimentally by checking the magnetic field produced by a very long rectilinear conductor. Before approaching the problem, two cases of special interest must be highlighted 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 0and the scalar product is simply the product of the magnitudes *B.ds* .

-The second occurs if **B** and d **s** are perpendicular, in which case the scalar product is 0, since the angle between the vectors is 90º, whose cosine is 0.

Another important detail is the choice of curve C on which the field circulation is evaluated. Ampère’s law does not specify what it can be, but it must involve current distribution. It also does not say in which direction to go around the curve and there are two possibilities for this.

The solution is to assign signs according to the rule of the right thumb. The four fingers are curved in the direction in which you want to integrate, usually this will be the same in which the **B** field circulates. 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 can be positive and some negative. The algebraic sum of them is the one that we are going to place in Ampère’s law, which is usually called an *enclosed current* (by curve C).

**Infinite rectilinear wire magnetic field**

In Figure 2 a wire is shown carrying a current I out of plane. The right thumb rule ensures that **B** circulates counterclockwise, describing circumferences as shown by the red arrows.

Let’s take one of them, whose radius is r. We divide it into small differential segments d **s** , represented by the vectors in blue. Both vectors, **B** and d **s** , are parallel at each point on the circumference, and in this way the integral ∫ _{C } **B** ∙ d **s** becomes:

∫ _{C} Bds

This is due to the fact that, as we said before, the scalar product **B** ∙ d **s** is the product of the magnitudes of the vectors and the cosine of 0º. We know the result of the integral thanks to Ampère’s law, therefore we write:

∫ _{C} Bds = μ _{or} I

Since the magnitude of the field is constant over the entire trajectory, it leaves the integral:

B ∫ _{C} ds = μ _{or} I

The integral ∫ _{C} ds represents the sum of all the infinitesimal segments that make up the circumference of radius *r* , equivalent to its length, the product of its radius by 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 ampereian circuit* ) were not circular, __the result__ of the integral continues to be μ _{or} I, however ∫ _{C } **B** ∙ d **s would **__no longer be__ B.2πr.

That is why the usefulness of Ampère’s law to determine the magnetic field lies in choosing distributions with high symmetry, such that the integral is easy to evaluate. Circular and rectilinear paths meet this requirement.

__Solved exercises__

__Solved exercises__

**– Exercise 1**

Consider the curves a, b, c and d shown in Figure 3. They involve three currents, two leaving the plane, symbolized by a point ( **.** ), Whose intensities are 1 A and 5 A, and one current entering the plane, which is denoted by a cross and whose magnitude is 2 A.

Find the current enclosed by each curve.

**Solution**

The currents coming out of the paper are assigned a + sign. According to this:

**Curve a**

It encloses the three currents, therefore the enclosed current is + 1 A + 5 A – 2 A = 4 A.

**Curve b**

Only the 1 A and – 2 A currents are within this curve, therefore the enclosed current is – 2 A.

**Curve c**

It encloses the outgoing currents 1A and 5 A, therefore the current locked up is 6 A.

**Curve d**

The currents inside it 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 located 1 meter from it, if the wire carries a current of 1 A.

**Solution**

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.

**References**

- Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 6. Electromagnetism. Edited by Douglas Figueroa (USB).
- Knight, R. 2017. Physics for Scientists and Engineering: a Strategy Approach. Pearson.
- Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 2.
- Serway, R. 2009. College Physics. Cengage Learning.
- Tipler, P. (2006) Physics for Science and Technology. 5th Ed. Volume 2. Editorial Reverté.