Question

What is the magnetic field $\overrightarrow{dB}$ at a distance $\overrightarrow{r}$ due to a small current element $\overrightarrow{dl}$ carrying current I?

• A. $\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }i\left(\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r}\right)$
• B. $\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }{i}^2\left(\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r}\right)$
• C. $\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }{i}^2\left(\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r^2}\right)$
• D. $\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }i\left(\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r^3}\right)$

Answer 1

The correct option is D $\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }i\left(\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r^3}\right)$

$dB=\frac{{\mu }_0}{4\pi }.\frac{idlsin\theta }{r^2}\Rightarrow \overrightarrow{dB}=\frac{{\mu }_0}{4\pi }.\frac{i(\overrightarrow{dl}\times \overrightarrow{r})}{r^3}$