question_answer

If an electron moves in a circular path of radium 15 cm in a magnetic Held of \[4\times {{10}^{-4}}T\], the speed of electron will be :

A)  \[5.87\times {{10}^{8}}m/s\]

B)  \[3.01\times {{10}^{4}}m/s\]

C)  \[2.03\times {{10}^{6}}m/s\]

D)  \[1.06\times {{10}^{7}}m/s\]

Correct Answer: D

Solution :

Key Idea: Magnetic force provides, the necessary centripetal force to the electron. The magnetic force acting on the particle is \[F=qvB\] ... (i) where q is charge, v is velocity and B is magnetic field. The centripetal force is \[qvB=\frac{m{{v}^{2}}}{r}\] ... (ii) where v is velocity and r is radius of circular path. Equating Eqs. (i) and (ii), we get \[qvB=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \] \[v=\frac{qBr}{m}\] Given, \[q=1.6\times {{10}^{-19}}C,\,r=15\,cm\] \[=15\times {{10}^{-2}}m\] \[B=4\times {{10}^{-4}}T,\,m=9\times {{10}^{-31}}kg\] \[\therefore \] \[v=\frac{15\times {{10}^{-2}}\times 4\times {{10}^{-4}}\times 1.6\times {{10}^{-19}}}{9\times {{10}^{-31}}}\] \[v=1.06\times {{10}^{7}}m/s\] Note: If the speed of electron increases its radius also increases, so that the time taken to complete one revolution remains same.