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An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall
$B = \frac{{2bm{v_0}}}{{\left( {{b^2} - {a^2}} \right)e}}$
$B = \frac{{2bm{v_0}}}{{\left( {{a^2} + {b^2}} \right)e}}$
$B = \frac{{m{v_0}}}{{\left( {\sqrt {{b^2} - {a^2}} } \right)e}}$
$B = \frac{{2m{v_0}}}{{\left( {\sqrt {{b^2} - {a^2}} } \right)e}}$
Solution
Pythagores theorem
$\mathrm{a}^{2}+\mathrm{R}^{2} =(\mathrm{b}-\mathrm{R})^{2}$
$=\mathrm{b}^{2}+\mathrm{R}^{2}-2 \mathrm{bR}$
so, $\frac{\mathrm{mv}_{0}}{\mathrm{qB}}=\mathrm{R}=\left(\frac{\mathrm{b}^{2}-\mathrm{a}^{2}}{2 \mathrm{b}}\right)$
