A) \[7.5m\]
B) \[7.5\times {{10}^{-2}}m\]
C) \[7.5\times {{10}^{-4}}m\]
D) \[7.5\times {{10}^{-3}}m\]
Correct Answer: C
Solution :
The velocity acquired by the electron is given as follows : \[\frac{1}{2}m{{v}^{2}}=eV\Rightarrow v=\sqrt{\frac{2eV}{m}}\] \[\Rightarrow \]\[v=\sqrt{\frac{2\times 1.6\times {{10}^{-19}}\times 500}{9.1\times {{10}^{-31}}}}=1.33\times {{10}^{7}}m/s\] when the particle enters the magnetic field its speed remains unchanged \[\therefore \]\[\frac{m{{v}^{2}}}{r}=qvB\Rightarrow r=\frac{mv}{qB}\] \[r=\frac{9.1\times {{10}^{-31}}\times 1.33\times {{10}^{7}}}{1.6\times {{10}^{-19}}\times 100\times {{10}^{-3}}}=7.56\times {{10}^{-4}}m\]
You need to login to perform this action.
You will be redirected in
3 sec
