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(A) The electron moves in a circular path of radius $R$ in the magnetic field,given by $R = \frac{mv_0}{eB}$.From the geometry of the problem,the cent

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$B = \frac{2bmv_0}{(b^2 - a^2)e}$

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(A) The electron moves in a circular path of radius $R$ in the magnetic field,given by $R = \frac{mv_0}{eB}$.From the geometry of the problem,the centre of the circular path lies on the $x$-axis at a distance $R$ from the wall.The distance from the centre of the sphere to the centre of the circular path is $(b - R)$.In the right-angled triangle formed by the radius of the sphere $a$,the radius of the circular path $R$,and the distance $(b - R)$,we have:$a^2 + R^2 = (b - R)^2$$a^2 + R^2 = b^2 + R^2 - 2bR$$a^2 = b^2 - 2bR$$2bR = b^2 - a^2$$R = \frac{b^2 - a^2}{2b}$Substituting $R = \frac{mv_0}{eB}$ into the equation:$\frac{mv_0}{eB} = \frac{b^2 - a^2}{2b}$$B = \frac{2bmv_0}{(b^2 - a^2)e}$

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