Year 1 Sem 2/LC Electromagnetism I/Lecture 15

Lecture 15: Magnitism IV: Ampere's Law
Mon 09 Mar 2026 11:00

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Biot-Savat: Complex Loops Continued

Continuing from the end of last lecture, we found by symmetry that the $yz$ components cancel, leaving only an $x$ component for any point $P$ along the x-axis.

Taking magnitudes (and noting that $\hat{r}$ is a unit vector):

$$\delta B = \frac{\mu_0 I}{4 \pi r^2} \delta l$$

Resolving into the x-direction, we get that:

$$\delta B_{x} = \frac{\mu_0 I}{4 \pi r^2} \delta l \cos \theta \tag{1}$$

And as we have a right angled triangle:

$$r^2 + a^2 + x^2 \tag{2}$$

$$\cos \theta = \frac{a}{r} \implies \cos \theta = \frac{a}{\left(a^2 + x^2\right)^{(1/2)}} \tag{3}$$

Inserting 2 and 3 into 1:

$$\delta B_x = \frac{\mu_0 I}{4 \pi} \times \frac{a}{\left(a^2 + x^2\right)^{(3/2)}} \delta l$$

$$B_x = \frac{\mu_0 I}{4 \pi} \times \frac{a}{\left(a^2 + x^2\right)^{(3/2)}} \oint \delta l$$

$$B_x = \frac{\mu_0 I}{4 \pi} \times \frac{a}{\left(a^2 + x^2\right)^{(3/2)}} 2 \pi a$$

$$B_x = \frac{\mu_0 I}{2} \frac{a^2}{\left(a^2 + x^2\right)^{(3/2)}}$$

Taking limits, we can observe that at the centre of the loop (where $x=0$):

$$B_0 = \frac{\mu_0 I}{2} \frac{a^2}{\left(a^2\right)^{(3/2)}}= \frac{\mu_0 I}{2a}$$

And when $x\gg a$, we have:

$$B_x = \frac{\mu_0 I a^2 }{2x^3}$$

However, we can simplify using the magnetic dipole moment $\mu_m = I \times \text{area} = I \pi a^2$, hence:

$$B_x = \frac{\mu_0}{4 \pi} \frac{2 \mu_m}{x^3}$$

This is strikingly similar for the expression for an electric dipole:

$$E_x = \frac{1}{4 \pi \epsilon_0} \frac{2p}{x^3}$$

E- and B-Fields

We have already determined the following, for E- and B-fields respectively:

$$\int_S \underline{E} \cdot d \underline{S} = \frac{Q}{\epsilon_0} \tag{E}$$

$$\int_S \underline{B} \cdot d \underline{S} = 0 \tag{B}$$

We also know that the E-field is conservative, hence no work is done if a charge returns to its original position:

$$\oint \underline{E} \cdot d \underline{l} = 0$$

Is there an equivalent of this for B-fields?

Consider a circiular path of a B-field created around an infinite line of current at radial distance $r$. By symmetry, $\underline{B}$ and $d \underline{l}$ are parallel, and are constant for a fixed $r$. Hence:

$$\oint \underline{B} \cdot d \underline{l} = \oint B dl = B \oint dL = B 2 \pi r$$

We have also determined that:

$$B = \frac{\mu_0 I}{2 \pi r} \implies \oint \underline{B} \cdot d \underline{l} = B 2 \pi r = \frac{\mu_0 I}{2 \pi r} \times 2 \pi r = \mu_0 I$$

This gives us Ampere’s Law:

$$\boxed{\oint \underline{B} \cdot d \underline{l} = \mu_0 I_\text{enclosed}}$$

Example

Consider a long solid cylinder carrying a uniformly distributed current. The cylinder has radius $R$ and length

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