
$\mathrm{a}^{2}+\mathrm{R}^{2} =(\mathrm{b}-\mathrm{R})^{2}$
$=\mathrm{b}^{2}+\mathrm{R}^{2}-2 \mathrm{bR}$
so, $\frac{\mathrm{mv}_{0}}{\mathrm{qB}}=\mathrm{R}=\left(\frac{\mathrm{b}^{2}-\mathrm{a}^{2}}{2 \mathrm{b}}\right)$


$\vec E = 2\hat i + 3\hat j ;\, B = 4\hat j + 6\hat k$
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