A) \[15\times {{10}^{6}}m/s\]
B) \[25\times {{10}^{6}}m/s\]
C) \[5\times {{10}^{6}}\text{ }m/s\]
D) \[1\times {{10}^{6}}\text{ }m/s\]
Correct Answer: C
Solution :
In Thomsons set up, for no deflection of electron. Force due to magnetic field = force due to electric field \[qvB=qE\] where q is quantity of change v the velocity, B the magnetic field and E the electric field. \[\therefore \] \[v=\frac{E}{B}\] Given, \[E=30\text{ }V\text{ }c{{m}^{-1}}=30\times 100\text{ }Vm\] \[B=6G=6\times {{10}^{-4}}T\] \[\therefore \] \[v=\frac{30\times 100}{6\times {{10}^{-4}}}=5\times {{10}^{6}}m/s\]
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