Question

A point charge of magnitude q = 4.5 nC is moving with speed v = $3.6\times {10}^7$ m/s parallel to the x-axis along the line y = 3 m. Find the magnetic field at the origin produced by this charge when the charge is at the point x = -4 m, y = 3 m, as shown in the figure.

• A. $-3.89\times {10}^{-10}\hat{k}T$
• B. $3.89\times {10}^{-10}\hat{k}T$
• C. $3.89\times {10}^{-10}\hat{i}T$
• D. $3.89\times {10}^{-10}\hat{j}T$

Answer 1

The correct option is A $-3.89\times {10}^{-10}\hat{k}T$

The magnetic field is given by $\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \hat{r}}{r^2},with\overrightarrow{v}=v\hat{i}$
$\overrightarrow{r}=(4\hat{i}-3\hat{j})\Rightarrow r=\sqrt{4^2+3^2}=5m$

Unit vector in the direction of $\hat{r}=\frac{\overrightarrow{r}}{r}=\frac{4\hat{i}-3\hat{j}}{5}=0.8\hat{i}-0.6\hat{j}$
$\overrightarrow{v}\times \hat{r}=\left(v\hat{i}\right)\times (0.8\hat{i}-0.6\hat{j})=-0.6v\hat{k}$

$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \hat{r}}{r^2}=\frac{{\mu }_0}{4\pi }\frac{q(-0.6v\hat{k})}{r^2}=-\left({10}^{-7}\right)\frac{(4.5\times {10}^{-9})\left(0.6\right)(3.6\times {10}^7)}{5^2}\hat{k}=-3.89\times {10}^{-10}T\hat{k}$