Question

A charged particle A of charge q = 2 C has velocity v = 100 m/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 = 2 m).the agle between them is 30.

Answer 1

$\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \overrightarrow{r}}{r^3}=\frac{{\mu }_0}{4\pi }\frac{qvsin\theta }{r^2}$

=$\frac{{\mu }_0}{4\pi }\frac{2\times 100\times sin{30}^0}{4}$

=${10}^{-7}\times 25$