Question

A charged particle $A$ of charge $q=2C$ has velocity $v=100m/s$. When it passes through point $A$ and has velocity in the direction shown, the strength of magnetic field at point $B$ due to this moving charge is $(r=2m)$

• A. $2.5\mu T$
• B. $5.0\mu T$
• C. $2.0\mu T$
• D. $noneofthese$

Answer 1

The correct option is A $2.5\mu T$

Magnetic field using Biot Savart law
$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }i\frac{(\overrightarrow{dl}\times \overrightarrow{r})}{{\overrightarrow{r}}^3}$
Though the expression above has current $i$, it can be recast into
$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }q\frac{(\overrightarrow{v}\times \overrightarrow{r})}{{\overrightarrow{r}}^3}$
$\therefore \left|\overrightarrow{B}\right|=\left({10}^{-7}\right)\times 2\times \frac{100\times \left(\sin \theta \right)}{2^2}$
$\therefore \left|\overrightarrow{B}\right|=\left({10}^{-7}\right)\times 2\times \frac{100\times \sin {30}^o}{4}$
$\therefore \left|\overrightarrow{B}\right|=2.5\times {10}^{-6}T=2.5\mu T$