Question

Expression of magnetic field using Biot- Savart law for a charge $q$ moving with velocity $v$ as shown in the figure is,

• A. $\overrightarrow{B}=\frac{{\mu }_0{q}^2}{4\pi }\frac{\overrightarrow{v}\times \overrightarrow{r}}{r^3}$
• B. $\overrightarrow{B}=\frac{{\mu }_{0}q}{2\pi }\frac{\overrightarrow{v}\times \overrightarrow{r}}{r^3}$
• C. $\overrightarrow{B}=\frac{{\mu }_{0}q}{4\pi }\frac{\overrightarrow{v}\cdot \overrightarrow{r}}{r^3}$
• D. $\overrightarrow{B}=\frac{{\mu }_{0}q}{4\pi }\frac{\overrightarrow{v}\times \overrightarrow{r}}{r^3}$

Answer 1

The correct option is D $\overrightarrow{B}=\frac{{\mu }_{0}q}{4\pi }\frac{\overrightarrow{v}\times \overrightarrow{r}}{r^3}$

Using Biot Savart's law,

$\left|\overrightarrow{dB}\right|=\frac{{\mu }_{0}i}{4\pi }\frac{dl\sin \theta }{r^2}$

Now multiplying and dividing by small time interval $dt$ in numerator and denominator,

$\left|\overrightarrow{dB}\right|=\frac{{\mu }_{0}i}{4\pi }\frac{dl\sin \theta }{r^2}\frac{dt}{dt}$

Now $\left(idt\right)$ can be written as $q$ because $i=\frac{q}{dt}=$ charge passing in unit time.

Similarly, $\frac{dl}{dt}$ can be written as $v$ as $v=\frac{dl}{dt}$.

$\therefore \left|\overrightarrow{dB}\right|=\frac{{\mu }_0}{4\pi }\frac{qv\sin \theta }{r^2}$

So, $\left|\overrightarrow{dB}\right|$ can be written as,

$\overrightarrow{dB}=\frac{q{\mu }_0}{4\pi }\frac{\overrightarrow{v}\times \overrightarrow{r}}{r^3}$

$\overrightarrow{\left|dB\right|}=\overrightarrow{B},$ because only one charge $q$ is present.

$\overrightarrow{B}=\frac{{\mu }_{0}q}{4\pi }\frac{\overrightarrow{v}\times \overrightarrow{r}}{r^3}$

Hence, option $\left(d\right)$ is correct.