# Ampere's Law

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16.6Ampere's Law

In this Physics tutorial, you will learn:

• What is the integral of a function? What is its geometrical meaning?
• How to find the integral of basic functions?
• What does Ampere's law say?
• What is an Amperian loop?
• What is the equivalent of Ampere's law in electricity?
• How to find the magnetic field outside a long straight wire with current?
• How to find the magnetic field inside a long straight wire with current?
• How to apply Ampere's law to find the magnetic field inside a solenoid?
• What is a toroid and how to apply Ampere's law to find the magnetic field inside the toroid?

## Introduction

So far, we have discussed more than once about the interaction between electric current and magnetic field. Such an interaction produces a number of physical quantities such as magnetic force, torque, motor effect (motion), magnetic moment, dipole moment, work, energy and so on.

There are many other quantities and relationships we can find using the similarities between magnetic and electric properties of matter. One of them is Ampere's Law, which is a mathematical statement of the relationship between currents and the magnetic fields they generate. It is thus the magnetic equivalent of Gauss's law, which relates charges to their electric fields. In this tutorial, we will discuss extensively about such a relationship, providing some mathematical background required to understand better the material discussed.

## Useful Mathematical Background: Integral of a function - Geometrical approach

When the graph of a function has a non-complex shape - such as a straight line, parabola, trapezium and so on - the area under the graph is easy to calculate as we use known geometric formulae to find it. Look at a few examples below. However, for more complex curves we cannot find the area under the graph that easily. In such situations, we divide the area under the graph in very small vertical strips of the same thickness Δx (more strips we have, better is) as shown in the figure below. Each strip can be considered as a small trapezium of height Δx the bases of which have a difference in length of Δy. If we choose to consider one of the small trapeziums (the i-th trapezium), we obtain for its area based on the known formula:

Area of trapezium = (Longer base + Shorter base) ∙ Height/2
Ai = (yi + yi-1) ∙ ∆x/2

Since

yi + yi-1/2 = < yi >

where < yi > is the average value of function in the i-th interval, we obtain for the area of the i-th trapezium:

Ai = < yi > ∙ ∆x

When we want to calculate the total area under the graph, we consider all small trapeziums formed when dividing the area according the above way. Thus, we write

Atot = Ai = < yi > ∙ ∆x

Usually we write f(x) instead of y. Thus, we have for the area under the graph.

Atot = < f(xi )> ∙ ∆x
= < f(x1 )> ∙ ∆x + < f(x2 )> ∙ ∆x + ⋯ + < f(xi )> ∙ ∆x + ⋯ + < f(xn )> ∙ ∆x

where < f(x1) >, < f(x2) >, < f(xi) >, < f(xn) >, are the average values of function in the 1st, 2nd, ith and nth trapezium respectively. (We have used the method of dividing the graph in small trapeziums when calculating the instantaneous velocity for example. In that case, we considered the slope of the graph around a given point, which corresponds to the lateral side of the small trapezium considered).

The area under the graph calculated in the above way is not 100% accurate as the graph is often a curve, not a straight line. The accuracy increases when we increase the number of divisions (the number of trapeziums therefore) so that the curve resembles more and more to a straight line (remember the Earth shape; it is curved but for small distances it looks flat). Therefore, the accuracy of the area-under-the-graph calculation would be very high if we increased the number of divisions (of small trapeziums) to infinity. In this case, we don't use anymore the symbol "Δ" to represent the width of interval but the symbol "d" instead, which is a symbol used for infinitely small intervals. Also, we use the symbol "" instead of "Σ" to represent the sum of all the individual areas of the small trapeziums. In this case, the average value of function fits more and more its lower and upper value in the given trapezium (trapeziums looks more and more as rectangles). Therefore, the last equation becomes

Atot = f(x)dx

The above expression is called the integral of the function f(x). Geometrically, it represents the area confined by the graph, the horizontal axis and the two vertical lines drawn from the extremities of the graph to the horizontal axis. The small horizontal segment "dx" is known as the differential part of the integral. It represents the width of each small vertical strip (interval). In other words, dx represents the height of each small trapezium obtained through the above method.

There is a specific method to find the value of integral for a given function. It is widely discussed in mathematics textbooks, but here we will give a few examples of integrals of some ordinary functions.

### 1- The integral of a constant:

a dx = a ∙ x

where a can be aby number, including 1.

xn dx = xn-1/n-1

ex dx = ex

### 4- The integral of sine and cosine function

sin⁡x dx = -cos⁡x
and
cos⁡x dx = sin⁡x

1/x dx = ln⁡x

### 6- The integral of differential gives the function itself as they are the inverse of each other. For example:

d (x2 ) = 2x dx
= 2x2/2
= x2

Some of the above integrals are used to describe magnetic properties of matter such as Ampere's law, which we will explain in the next paragraph.

## Ampere's Law

In the tutorial 14.6 "Electric Flux - Gauss Law", we explained that the net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. For example, the net outward electric flux through a charged sphere is

Φ = E ∙ A
= 1/4πϵ0Q/R2 ∙ 4πR2
= Q/ϵ0

We can use the integral method to find the same result. Thus, for a small segment dA of the sphere, we have

dΦ = E ∙ dA

The total flux through the entire surface A is calculated by taking the integral of the above expression. We have:

Φ = d Φ
= E ∙ dA
= E ∙ A
= 1/4πϵ0Q/R2 ∙ 4πR2
= Q/ϵ0

As you see, the result obtained is the same for both methods used.

Similarly, we can find the net magnetic field due to any distribution of currents by first considering the differential magnetic field dB. The magnitude of the elementary magnetic field produced by a current-length element idL at the given point P, which is at a distance r from the given current element, is (the scalar version):

dB = μ0/4nidL ∙ sinθ/r2 The above formula is known as the Biot-Savart Law. If the current distribution has some symmetry, we can integrate the above expression to obtain a simpler expression for the magnetic field. Thus, after making a few arrangements in the integral, we have

B dL = μ0 ∙ i

The above expression is known as the Ampere's Law and it is especially useful when considering the current flowing through a closed loop. In such cases, we can write:

B dL = μ0 ∙ iencl

By definition, Ampere's Law states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability of free space times the electric current enclosed in the loop.

The circle in the symbol of integral shows that it is the integral of a closed path (loop). Geometrically, it represents the area bordered by the graph only, not by the horizontal axis as in normal integrals, as shown in the figure. Any closed loop as those discussed above is known as Amperian loop. The figure below shows an Amperian loop, which encircles two current carrying wires but excludes a third one. The directions of currents are shown through the known symbols ⊗ and ⨀. The direction of the currents is important to determine the current signs in the final formula after integration. For this purpose, we use the curled right hand rule in which the four fingers are placed in the direction of integration and the outstretched thumb shows the direction of a positive current. If the current is in the negative direction, it is taken as negative. As for the direction of magnetic field B, regardless its direction, it is generally assumed in the direction of integration for simplicity. This means it is not necessary to know the direction of magnetic field prior to integration.

For example, in the figure above, i1 is positive as it is out of page and i2 is negative (onto the page). Also, we have introduced an angle θ in the figure as the direction of magnetic field B is not exactly the same as that of the loop element dL in the given position. Hence, Ampere's law in vector form becomes

B dL = B⋅cos⁡θ dL
= μ0 ∙ iencl

It is obvious that the expression BdL inside the integral represents the scalar product of an elementary element dL of an Amperian loop and the magnetic field B in that position, which is tangent to the element dL. This means the value of integral represents the sum of all such products around the entire loop.

In our figure, the net current is

inet = i1 - i2

Therefore, using Ampere's law for these two currents we obtain:

B dL = B⋅cos⁡θ dL
= μ0 ∙ iencl
= μ0 ∙ (i1 - i2)

For example, the value of integral above if i1 = 5A and i2 = 3 A is

B dL = μ0 ∙ (i1 - i2 )
= 4π × 10-7 N/A2 ∙ (5A - 3A)
= 8π × 10-7 N/A

Now, let's consider a couple of examples in which we can apply simple integration techniques to find what we are missing.

## Magnetic Field Outside a Long Straight Wire with Current

This situation is similar to the one discussed above. The only difference is that now we are considering a single current-carrying wire instead of two parallel or antiparallel ones. Therefore, the magnetic field B will have a cylindrical symmetry and any Ampere's loop we want to consider will have a circular shape, as shown in the figure. The current direction in this case is out of page. We enclose the current inside the desired Amperian loop of radius r, which is concentric to the wire's cross section. In this way, the integral (which is in the counter-clockwise direction) takes a simpler form than if the loop had a more complicated shape.

Giving that the magnetic field B is tangent to the loop, its vector is either parallel or antiparallel to the Amperian loop element dL. For simplicity, it is better to assume them as parallel (θ = 0), so that cos θ = 1. Therefore, we obtain for Ampere's law in this situation:

B dL = B⋅cos⁡θ dL
= B ∙ dL
= B ∙ (2π ∙ r)
= μ0 ∙ iencl

Considering the last two identities

B ∙ (2π ∙ r) = μ0 ∙ iencl

we obtain the well-known formula of magnetic induction (in scalar form) produced by a long current carrying wire at distance r from the wire, which we have provided in tutorial 16.2:

B = μ0 ∙ iencl/2π ∙ r

### Example 1

A magnetic field of magnitude 5mT is measured at a distance of 4cm from a long straight wire as shown in the figure. #### Calculate:

1. The direction of electric current flowing through the wire
2. The magnitude of this current

### Solution 1

Clues:

r = 4 cm = 4 × 10-2 m
B = 0.5 mT = 5 × 10-4 T
0 = 4π × 10-7 N/A2)

1. The direction of current is found using the right hand rule. We grasp the wire by the right hand where the curled fingers show the direction of magnetic field. The outstretching thumb therefore shows the direction of current. In this case, this direction is from up to down.
2. Using the equation obtained by applying Ampere's law for the magnetic field outside a long-straight wire,
B = μ0 ∙ iencl/2π ∙ r
we obtain for the electric current flowing through the wire
iencl = B ∙ 2π ∙ r/μ0
= (5 × 10-4 T) ∙ (2π) ∙ (4 × 10-2 m)/(4π × 10-7 N/A2 )
= 102 A
= 100A

## Magnetic Field Inside a Long Straight Wire with Current

In this case, the radius r of Amperian loop is smaller than the radius R of wire. This means the Amperian loop is taken inside the wire's section. Like in the previous paragraph, the distribution of magnetic field has a cylindrical symmetry as the current is uniformly distributed throughout the wire (the direction is out of page). Therefore, we use a similar approach as in the calculation of magnetic field outside the wire discussed in the previous paragraph.

Using Ampere's law for this situation, we obtain for r < R:

B dL = B ⋅ cos⁡θ dL
= B ∙ dL
= B ∙ (2π ∙ r)
= μ0 ∙ iencl

Since the current distribution is uniform, the current enclosed in the loop is proportional to the area of loop at any point, i.e.

iencl/itot = π ∙ r2/π ∙ R2

or

iencl = itotπ ∙ r2/π ∙ R2

Substituting this value for current in the previous formula obtained after integrating Ampere's law equation, yields

B ∙ (2π ∙ r) = μ0 ∙ itotπ ∙ r2/π ∙ R2

or

B = μ0 ∙ itot/2π ∙ R2 ∙ r

The above formula means the magnetic field is zero at centre of cross section of the wire and it has a maximum value for r = R (on the outer surface of wire). This is another proof that the electric charges (that are responsible for the magnetic field generation) are concentrated on the outer layer of conductor, as we have explained in Section 14.

## Ampere's Law Applied in Solenoids and Toroids

In tutorial 16.2 we have explained that the magnetic field produced by a solenoid which contains N turns is

B = μ0 ∙ N ∙ I/L

If we express this equation in terms of the number of turns per unit length n instead of N/L, we obtain for the magnetic field inside the solenoid

B = μ0 ∙ n ∙ I

This formula is obtained even when we use Ampere's law approach. For this, we have to consider a rectangular Amperial loop abcd as shown below. An ideal solenoid has a uniform magnetic field inside (the parallel lines in the figure) and zero outside it. Using the rectangular loop abcda, we write the closed integral BdL as the sum of four integrals (one for each segment). Thus, we have:

BdL = baBdL + cbBdL + dcBdL + adBdL

The only integral, which has a non-zero value, is the first one. On the other hand, the integral according the path cd is zero because magnetic field outside the solenoid is zero. Also, the integrals according the paths bc and da are both zero as the magnetic field lines are perpendicular to the paths (cos 90° = 0). Therefore, if we denote the length ab = h we obtain

BdL = B ∙ h

As for the net current Iencl enclosed within the Amperian loop, we have

iencl = I ∙ (n ∙ h)

where 'I' is the current flowing through the entire solenoid.

Combining the general form of Ampere's law

BdL = μ0 ∙ iencl

with the above (transformed) form of this law, we obtain

B ∙ h = μ0 ∙ I ∙ (n ∙ h)

or

B = μ0 ∙ n ∙ I

This is the same formula obtained in 16.2 for the magnetic field inside a solenoid but now using Ampere's law.

A toroid is a hollow solenoid that has been curved until its two ends meet, forming a sort of hollow bracelet. We can determine the magnetic field inside the toroid (inside the bracelet-like shape) using Ampere's law.

From symmetry of toroid it is easy to conclude that the magnetic field lines inside a toroid are concentric circles. We can choose a concentric circle of radius r as an Amperian loop. When solving the integral of Ampere's Law for toroid

BdL = μ0 ∙ iencl

we obtain

B ∙ 2π ∙ r = μ0 ∙ iencl ∙ N

where i is the current in the toroid windings (it is positive for windings included inside the Amperian loop) and N is the total number of turns. Hence, we obtain for the magnetic field inside the toroid:

B = μ0 ∙ iencl ∙ N/2π ∙ r

Unlike in solenoids, the magnetic field B is not constant throughout the cross-section of toroids. In addition, the magnetic field outside a toroid is zero.

The direction of magnetic field in toroid can be determined in the same way as in solenoids, i.e. by using the curled right-hand rule in which we grasp the toroid with our right hand (the four fingers are in the direction of current) and the outstretched thumb shows the direction of magnetic field.

### Example 2

A toroid having 500 turns has an inner diameter d = 3 cm and an outer diameter D = 5 cm. What is the magnetic field at centre of cross section of toroid if the current flowing through it is 20 mA?

### Solution 2

The centre of cross-section of toroid is at any midpoint between the inner and outer radius. This position corresponds to the value of r in the formula (as in the figure shown in theory). Thus, we have

r = d + D/2
= 3 cm + 5cm/2
= 4 cm
= 4 × 10-2 m

Giving that N = 500 = 5 × 102 and iencl = 20 mA = 0.02 A = 2 × 10-2 A, we obtain

B = μ0 ∙ iencl ∙ N/2π ∙ r
= (4π × 10-7 N/A2 ) ∙ (2 × 10-2 A) ∙ (5 × 102 )/2π ∙ (4 × 10-2 m)
= 5 × 10-5 T

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