Respuesta :
Answer:
The electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]
The magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]
Explanation:
- Calculating the electric field at the observation point:
We are given:
[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m[/tex]
The equation used to calculate the electric field follows:
[tex]\vec{E}=\frac{kq\hat{r}}{r^2}[/tex]
OR
[tex]\vec{E}=\frac{kq}{r^2}\frac{\vec{r}}{|\vec{r}|}[/tex]
We know:
[tex]|\vec{r}|=r[/tex]
[tex]q=1.6\times 10^{-19}C[/tex]
So, the equation becomes:
[tex]\vec{E}=\frac{kq(\vec{r})}{r^3}[/tex] .....(1)
Putting values in equation 1, we get:
[tex]\vec{E}=\frac{(9\times 10^9}(1.6\times 10^{-18})(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})}{\left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{E}=(1.84\times 10^{18}(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})\\\\\vec{E}=1.65\times 10^{9}\hat{i}+3.68\times 10^{8}\hat{j}[/tex]
Hence, the electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]
- Calculating the magnetic field at the observation point:
The magnetic field due to moving charge is given by:
[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \hat{r}}{r^2}[/tex]
OR
[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \vec{r}}{r^3}[/tex] .....(2)
We are given:
[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m\\\\\vec{v}=-3\times 10^6\hat{i}[/tex]
Putting values in equation 2, we get:
[tex]\vec{B}=\frac{\mu_o\times (1.6\times 10^{-19})\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]}{4\pi\times \left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{B}=(20.42)\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]\\\\\vec{B}=-1.23\times 10^{-14}\hat{k}[/tex]
Hence, the magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]
