Ampere's Law

Ampere's Law Formula

In this section, we will explain Ampere's original form of the law, as well as Maxwell's addition, understanding the motivations behind both versions.

Ampere's Law

The equation for Ampere's law is given by

\[ \oint \vec{B} \cdot \mathrm{d} \vec{l} = \mu_0 I_{\text{enclosed}} ,\]

where \(\vec{B}\) is the magnetic vector field around the electric current measured in \(\mathrm{\frac{A}{m}}\), \(\mathrm{d} \vec{l}\) is the infinitesimal line element vector measured in \(\mathrm{m}\), \(\mu_0\) is the vacuum permeability given by a value of \(4\pi \times 10^{-7} \, \mathrm{\frac{H}{m}} \), and \( I_{\text{enclosed}}\) is the current enclosed by the amperian loop measured in amperes \(\mathrm{A}\). You should be familiar with this from previous calculus courses, but as a reminder, \(\oint\) represents the closed-loop one-dimensional line integral.

Ampere-Maxwell Law

Now let's go through the correction that Maxwell contributed to Ampere's law. One of the primary rules in physics that all equations must obey is conservation. In systems involving moving parts, all objects must obey energy conservation and momentum conservation. Similarly, in electric and magnetic systems, the system must obey the conservation of charge. Maxwell found that Ampere's law on its own does not obey the conservation of charge, therefore, there must be a missing term in this equation.

The new law that Maxwell modified is then given as

\[ \oint \vec{B} \cdot \mathrm{d} \vec{l} = \mu_0 I_{\text{enclosed}} + \mu_0 \epsilon_0 \frac{\mathrm{d} \Phi_{\text{E}}}{\mathrm{d} t} ,\]

where the new extra term includes vacuum permittivity \( \epsilon_0\) given by a value of \( 8.85 \times 10^{-12} \, \mathrm{\frac{F}{m}} \), \(\Phi_{\text{E}}\) is the flux of the electric field measured in \(\mathrm{V \, m} \), and \(t\) is the time variable measured in \(\mathrm{s}\).

Biot Savart Law vs Ampere's Law

Another equation that allows us to calculate the magnetic field generated by an electric current is the Biot Savart law. It provides us with a way of calculating the magnetic field given a more complex situation with less symmetry. The equation is given as

\[ \vec{B}(r) = \frac{\mu_0}{4 \pi} \int_C \frac{I \, \mathrm{d} \vec{l} \times \vec{r} }{| \vec{r} |^3} ,\]

where our integral now contains the infinitesimal line element \(\mathrm{d} \vec{l}\) along the integral path \(C\) and \(\vec{r}\) is the vector distance between the point on the wire to the point where the magnetic field is being computed.

An important point to note about this equation is the cross product between \(\mathrm{d} \vec{l}\) and \(\vec{r}\), this emphasizes to us that the direction of the resultant magnetic field is orthogonal to that of the current element.

The Biot Savart law grants us more flexibility when calculating the magnetic field due to a current-carrying wire. When using Ampere's law, we are bounded to situations where we can exploit the symmetry of the magnetic field, for instance, straight current-carrying wires, solenoids, or conducting slabs. In theory, the Biot Savart law could be applied to a current of any shape, as long as the integral is solvable.

Ampere's Law Examples

Now let's consider some questions to which we can apply Ampere's law.

We can also consider a second example where we find the magnetic field inside a current-carrying wire.

Ampere's Law Application

So far in the article, we have only covered the applications of Ampere's law in straight, current-carrying wires, but Ampere's law can also be applied in other current-carrying shapes, such as a solenoid. In previous discussions of electromagnetism, we have already come across the equation for the magnetic field inside a solenoid as

\[ B = \mu_0 \frac{N I}{L} ,\]

where \(N\) is the number of turns in the solenoid coil, \(L\) is the length of the coil measured in \(\mathrm{m}\), and all the other symbols are the same as per previous equations. In order to derive this equation using Ampere's law, let's consider a square amperian loop containing one side of the solenoid coil.

In order to solve for the magnetic field within the solenoid, we need to calculate the line integral. Referring to the figure, we can break up the line integral into four components representing the four edges of the rectangular loop. This results in

\[ \oint \vec{B} \cdot \mathrm{d} \vec{l} = \int_A^B \vec{B} \cdot \mathrm{d} \vec{l} + \int_B^C \vec{B} \cdot \mathrm{d} \vec{l} + \int_C^D \vec{B} \cdot \mathrm{d} \vec{l} + \int_D^A \vec{B} \cdot \mathrm{d} \vec{l} ,\]

where \(A, B, C\) and \(D\) represent the labeled edges of the rectangular loop. Now we can simplify and cancel out some components by analyzing the dot product. First considering the integral between edges \(B\) to \(C\) and \(D\) to \(A\), we can see that the magnetic field is perpendicular to the loop, rendering the dot product to zero. On the other hand, the integral between the point \(C\) to \(D\) is also zero due to the fact that the magnetic field just outside the solenoid is zero.

This leaves us with the final integral between the point \(A\) to \(B\); assuming that the section of the curve is of length \(L\), we can simplify it down to

\[ \oint \vec{B} \cdot \mathrm{d} \vec{l} = BL .\]

Finally, in order to combine this result with Ampere's law, we still need to calculate the amount of current enclosed by the loop. The current contained by \(N\) turns of the solenoid is given by

\[ I_{\text{enclosed}} = NI,\]

where \(I\) is the current in the coil and \(N\) is the number of enclosed turns by the amperian loop.

Thus we can plug all of these results into Ampere's law to find that

\[ \begin{align} \oint \vec{B} \cdot \mathrm{d} \vec{l} &= \mu_0 I_{\text{enclosed}} \\ BL &= \mu_0 NI \\ B &= \frac{\mu_0 NI}{L} . \end{align} \]