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, \[m=9\times {{10}^{-31}}kg\], \[e=1.6\times {{10}^{-19}}C\], \[v=4\times {{10}^{6}}m/s\], \[B=2\times {{10}^{-1}}T\] \[\therefore \] \[r=\frac{9\times {{10}^{-31}}\times 4\times {{10}^{6}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-1}}}\] \[=1.1\times {{10}^{-4}}m\]You need to login to perform this action.
You will be redirected in
3 sec