UY1: Magnetic Field Of A Straight Current Carrying Conductor

magnetic field of straight current carrying conductor


Consider a straight conductor with length 2a carrying a current I. Find the magnetic field at point P which is located at a distance x from the conductor on its perpendicular bisector.

From Biot-Savart Law, we have:

$$\vec{B} = \frac{\mu_{0}}{4 \pi} \int \frac{I}{r^{2}} \, d\vec{l} \times \hat{r}$$

First, we will need to work out what is $d\vec{l} \times \hat{r}$.

$$d\vec{l} = dl \, \hat{j}$$

$$\begin{aligned} \hat{r} &= \cos{\left( \phi-90^{\circ} \right)}\hat{i} – \sin{\left( \phi-90^{\circ}\right)}\hat{j} \\ &= \sin{\phi} \hat{i} + \cos{\phi} \hat{j} \end{aligned}$$


$$d\vec{l} \times \hat{r} = \, – \sin{\phi} \, dl \, \hat{k}$$

Substituting that into the Biot-Savart Law, we have:

$$\vec{B} = \left[ \frac{\mu_{0}I}{4 \pi} \int \frac{1}{r^{2}} \left( – \sin{\phi} \, dl \right) \right] \hat{k}$$

Notice that both $r^{2}$ and $\sin{\phi}$ can be replaced with:

$$r^{2} = x^{2} + y^{2}$$

$$\sin{\phi} = \frac{x}{\sqrt{x^{2}+y^{2}}}$$

We have:

$$\begin{aligned} \vec{B} &= \left[ \frac{\mu_{0}I}{4 \pi} \int\limits_{-a}^{a} \frac{1}{x^{2} + y^{2}} \left( – \frac{x}{\sqrt{x^{2}+y^{2}}} \, dy \right) \right] \hat{k} \\ &= -\frac{\mu_{0}I}{4 \pi} \int\limits_{-a}^{a} \frac{x \, dy}{\left(x^{2}+y^{2} \right)^{\frac{3}{2}}} \hat{k} \\ &= -\frac{\mu_{0}I}{4 \pi} \frac{2a}{x \sqrt{x^{2}+a^{2}}} \hat{k} \end{aligned}$$

The magnitude of the magnetic field is:

$$B = \frac{\mu_{0}I}{4 \pi} \frac{2a}{x \sqrt{x^{2}+a^{2}}}$$

What if the current carrying conductor is long (assumed to be infinite in length)?

From the above equation, we have:

$$\begin{aligned} B &= \frac{\mu_{0}I}{4 \pi} \frac{2a}{x \sqrt{x^{2}+a^{2}}} \\ &= \frac{\mu_{0}I}{2 \pi} \frac{1}{x \sqrt{1 + \frac{x^{2}}{a^{2}}}} \end{aligned}$$

For infinite length of wire, $a \rightarrow \infty$, $\frac{x^{2}}{a^{2}} \rightarrow \infty$. We have:

$$B = \frac{\mu_{0}I}{2 \pi x}$$

x is normally denoted by r, which is the distance from the conductor to the point.


$$B = \frac{\mu_{0}I}{2 \pi r}$$


Next: Magnetic Field & Force Between Parallel Conductors

Previous: Magnetic Field Of A Current Element

Back To Electromagnetism (UY1)

Back To University Year 1 Physics Notes

Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.