Question

A $-4.80\mu C$ charge is moving at a constant velocity of $6.80\times {10}^{5}m/s$ in the +x direction relative to a reference frame.At the instant when the point charge is at the origin,what is the magnetic field vector it produces at the following points x=0,y=0.500 m,z=0?

Answer 1

Here $q=-4.8\mu C$

Velocity of the point charge is $\overrightarrow{v}=6.8\times {10}^{5}m/s\hat{i}$

Magnetic field at position vector $\overrightarrow{r}=\frac{{\mu }_0}{4\pi }\frac{\left(q\right(\overrightarrow{v}\times \hat{r}\left)\right)}{r^2}$

For $y=0.5m$

$r=0.5\hat{j}$

$⟹B=\frac{{\mu }_0}{4\pi }\frac{(-4.8\times {10}^{-6})\left(\right(6.8\times {10}^5\hat{i})\times (\hat{j}\left)\right)}{{0.5}^2}$

$=-1.31\times {10}^{-6}T\hat{k}$