A) 2400 V/cm
B) 600 V/cm
C) 1200 V/cm
D) 12000 V/cm
Correct Answer: C
Solution :
Key Idea: The electric field acting on the particles provides the necessary centripetal force. Cathode rays are composed of electrons, when they move in electric field a force \[F=eE\] ... (i) acts on them, this provides the necessary centripetal force to the particles. \[\therefore \] \[F=\frac{m{{v}^{2}}}{r}\] ... (ii) From Eqs. (i) and (ii), we get \[eE=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \] \[r=\frac{m{{v}^{2}}}{eE}=\frac{m\,{{({{10}^{6}})}^{2}}}{e\,(300)}\] ?. (iii) When velocity is doubled same circular path is followed, hence radius is same \[r=\frac{m\,{{(2\times {{10}^{6}})}^{2}}}{eE}\] ?. (iv) Equating Eqs. (iii) and (iv), we get \[\frac{m\times {{({{10}^{6}})}^{2}}}{300\,e}=\frac{m\times {{(2\times {{10}^{6}})}^{2}}}{eE}\] \[\Rightarrow \] \[E=300\times 4=1200\,V/cm\] Note: Since, \[E\propto {{u}^{2}}\] and v is doubled much larger field will be required.
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