Question

A particle moving with velocity v= i+3j and it produces an electric field at a point given by e=2k

Answer 1

your complete question is ----> A particle is moving with velocity i^+3j^ and it produces an electric field 2k^ at a point. It will produce magnetic field at that point equal to...

1. (6i - 2j)/c² ,
2. (6i + 2j)/c² ,
3. cannot be determined
4. zero

solution : velocity of particle, $\vec{v}=\hat{i}+3\hat{j}$

electric field, $E=2\hat{k}$

we know,

electric field , E = $2\hat{k}=\frac{q}{4\pi\epsilon_0}\frac{\vec{r}}{|r|^3}$....(1)

magnetic field , $B=\frac{\mu_0}{4\pi}q\frac{\vec{V}\times\vec{r}}{|r|^3}$......(2)

[ here it is clear that magnetic field is perpendicular upon velocity of particle. so, magnetic field must be in such a way that dot product of it and its velocity = 0]

and relation between $\mu_0$ and $\epsilon_0$ is $\epsilon_0\mu_0=\frac{1}{c^2}$.....(3)

from equations (1) ,(2) and (3)

$B=\frac{\vec{V}\times\vec{E}}{c^2}$

$B=\frac{(\hat{i}+3\hat{j}\times2\hat{k}}{c^2}$

= (6i - 2j)/c²

hence, magnetic field of particle is (6i - 2j)/c²