question_answer

An electron having charge \[1.6\times {{10}^{-19}}C\]and mass \[\text{9}\times \text{1}{{0}^{-31}}\text{kg}\]is moving with \[\text{4}\times \text{1}{{0}^{\text{6}}}\text{ m}/\text{s}\] speed in a magnetic field of \[\text{2}\times \text{1}{{0}^{-\text{1}}}\] T in a circular orbit. The force acting on an electron and the radius of circular orbit will be

A)  \[1.28\times {{10}^{-14}}N,1.1\times {{10}^{-3}}m\]

B)  \[\text{1}.\text{28}\times \text{1}{{0}^{\text{15}}}\text{ N},\text{ 1}.\text{2}\times \text{1}{{0}^{-\text{12}}}\text{ m}\]

C)   \[\text{1}.\text{28}\times \text{l}{{0}^{-\text{13}}}\text{ N},\text{ 1}.\text{1}\times \text{l}{{0}^{-\text{4}}}\text{ m}\]

D)  None of the above

Correct Answer: C

Solution :

                 Electron moves in a magnetic field \[(\vec{B})\] in a circular orbit of radius (r), hence Centripetal force = force due to magnetic field \[\frac{m{{v}^{2}}}{r}=evB\] \[\Rightarrow \]               \[r=\frac{mv}{eB}\]                 Given,   \[m=9\times {{10}^{-31}}kg\],                    \[e=1.6\times {{10}^{-19}}C\],                 \[v=4\times {{10}^{6}}m/s\],                 \[B=2\times {{10}^{-1}}T\] \[\therefore \]  \[r=\frac{9\times {{10}^{-31}}\times 4\times {{10}^{6}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-1}}}\]                 \[=1.1\times {{10}^{-4}}m\]