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(A) The magnetic field $\overrightarrow{B}$ produced by a moving point charge is given by $\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{q(\overrighta

Vedclass · Accepted Answer

$\frac{6\hat{i} - 2\hat{j}}{c^2}$

Vedclass · Answer

(A) The magnetic field $\overrightarrow{B}$ produced by a moving point charge is given by $\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{q(\overrightarrow{v} 	imes \overrightarrow{r})}{r^3}$.The electric field $\overrightarrow{E}$ produced by the same charge at that point is $\overrightarrow{E} = \frac{1}{4\pi\epsilon_0} \frac{q\overrightarrow{r}}{r^3}$.By substituting the expression for $\frac{q\overrightarrow{r}}{r^3}$ from the electric field equation into the magnetic field equation,we get $\overrightarrow{B} = \epsilon_0 \mu_0 (\overrightarrow{v} 	imes \overrightarrow{E})$.Since $c^2 = \frac{1}{\mu_0 \epsilon_0}$,we have $\overrightarrow{B} = \frac{\overrightarrow{v} 	imes \overrightarrow{E}}{c^2}$.Given $\overrightarrow{v} = \hat{i} + 3\hat{j}$ and $\overrightarrow{E} = 2\hat{k}$,the cross product is:$\overrightarrow{v} 	imes \overrightarrow{E} = (\hat{i} + 3\hat{j}) 	imes 2\hat{k} = 2(\hat{i} 	imes \hat{k}) + 6(\hat{j} 	imes \hat{k}) = 2(-\hat{j}) + 6(\hat{i}) = 6\hat{i} - 2\hat{j}$.Therefore,$\overrightarrow{B} = \frac{6\hat{i} - 2\hat{j}}{c^2}$.

$A$ particle is moving with velocity $\overrightarrow{v} = \hat{i} + 3\hat{j}$ and it produces an electric field at a point given by $\overrightarrow{E} = 2\hat{k}$. It will produce a magnetic field at that point equal to (all quantities are in $SI$ units):

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