question_answer

In the question number 53, if velocity is normal in the shorter side then the voltage developed is:

A) \[2.3\times {{10}^{-4}}V\]       

B)             \[2.4\times {{10}^{-2}}V\]

C) \[4.8\times {{10}^{-2}}V\]                   

D) \[1.6\times {{10}^{-4}}V\]

Correct Answer: D

Solution :

(d) \[1.6\times {{10}^{-4}}V\] Here, \[l=2\,cm=2\times {{10}^{-2}}m\] \[\varepsilon =Blv=0.4\times 2\times {{10}^{-2}}\times 2\times {{10}^{-2}}\] \[=16\times {{10}^{-4}}V\]