• # question_answer An electron revolves in air in a circle in a plane perpendicular to the principal axis. The air has a uniform magnetic field parallel to the principal axis and of magnitude 2 tesla with centre of circle at a distance 10 cm from the pole of the spherical curved glass $(\mu =1.5)$ surface of radius of curvature 10 cm as shown in figure. An observer from P sees radius of circle to be 10 mm. The linear momentum of electron is A) $0.4\times {{10}^{-21}}kg-m/s$            B) $0.8\times {{10}^{-21}}kg-m/s$ C) $1.6\times {{10}^{-21}}kg-m/s$ D) $2.4\times {{10}^{-21}}kg-m/s$

[c]  Applying refraction formula for curved surface $\frac{{{\mu }_{2}}}{\upsilon }-\frac{{{\mu }_{1}}}{u}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$ $\Rightarrow$   $\frac{3}{2\upsilon }-\frac{1}{-10}=\frac{\frac{3}{2}-1}{+10}$ $\Rightarrow$   $\upsilon =-30\,cm$ Magnification $=\frac{{{h}_{i}}}{{{h}_{0}}}=\frac{{{\mu }_{1}}\upsilon }{{{\mu }_{2}}u}$
 $\Rightarrow$   ${{h}_{0}}=\frac{{{\mu }_{2}}u}{{{\mu }_{1}}\upsilon }{{h}_{i}}=\frac{\frac{3}{2}(-10)}{1\times (-30)}\times 10\,\,mm$ $\therefore$      ${{h}_{0}}=5\,\,mm=$ radius of circle of ${{e}^{-}}$ $\Rightarrow$   $5\,mm=\frac{(m\upsilon )}{eB}$  $\Rightarrow$ $m\upsilon =5\times {{10}^{-3}}\times 1.6\times {{10}^{-19}}\times 2\,\,kg-m/s$ $\therefore$      Momentum $=1.6\times {{10}^{-21}}\,kg-m/s$