
Figure: $Image$
$(A)$ The particle enters Region $III$ only if its velocity $V>\frac{q / B}{m}$
$(B)$ The particle enters Region $III$ only if its velocity $\mathrm{V}<\frac{\mathrm{q} / \mathrm{B}}{\mathrm{m}}$
$(C)$ Path length of the particle in Region $II$ is maximum when velocity $V=\frac{q / B}{m}$
$(D)$ Time spent in Region $II$ is same for any velocity $V$ as long as the particle returns to Region $I$
Due to the presence of the current $I_1$ at the origin


$\overrightarrow{\mathrm{F}} =\mathrm{q}(\vec{v} \times \overrightarrow{\mathrm{B}})$
$=\mathrm{q} \vec{v} \times\left(\mathrm{B} \hat{i}+\mathrm{B} \hat{j}+\mathrm{B}_{0} \hat{k}\right)$
For $\mathrm{q}=1$ and $\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}$ and
$\overrightarrow{\mathrm{F}}=4 \hat{i}-20 \hat{j}+12 \hat{k}$
What will be the complete expression for $\vec{B}$ ?
$\vec E = 2\hat i + 3\hat j ;\, B = 4\hat j + 6\hat k$
The charged particle is shifted from the origin to the point $P(x = 1 ;\, y = 1)$ along a straight path. The magnitude of the total work done is