An electron of mass m and charge e is project | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The electron moves in a circular path of radius $R$ in the magnetic field $B$. The radius of the path is given by $R = \frac{mv_0}{eB}$.From the geometry of the problem,consider the right-angled triangle formed by the centre of the sphere,the point of projection,and the centre of the circular path of the electron.The distance from the centre of the sphere to the centre of the circular path is $(b - R)$. The radius of the sphere is $a$,and the radius of the circular path is $R$. Since the electron is projected normally,the radius of the sphere at the point of projection is perpendicular to the velocity $v_0$,which is tangent to the circular path.Applying the Pythagorean theorem to the triangle with hypotenuse $(b - R)$,base $a$,and height $R$ is not correct here. Instead,the triangle has sides $a$,$R$,and hypotenuse $(b - R)$ is incorrect. Looking at the geometry,the distance from the centre of the sphere to the wall is $b$. The centre of the circular path lies on the $x$-axis. The distance from the centre of the sphere to the centre of the circular path is $(b - R)$.In the right triangle formed by the centre of the sphere,the point of projection,and the centre of the circular path,the hypotenuse is $(b - R)$,one leg is $a$,and the other leg is $R$. Thus,$(b - R)^2 = a^2 + R^2$.$b^2 + R^2 - 2bR = a^2 + R^2$$b^2 - a^2 = 2bR$$R = \frac{b^2 - a^2}{2b}$Substituting $R = \frac{mv_0}{eB}$,we get $\frac{mv_0}{eB} = \frac{b^2 - a^2}{2b}$.Solving for $B$,we get $B = \frac{2bmv_0}{(b^2 - a^2)e}$.

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