A) \[3.8\times {{10}^{-4}}V\]
B) \[4.8\times {{10}^{-4}}V\]
C) \[2.2\times {{10}^{-2}}V\]
D) \[3.2\times {{10}^{-2}}V\]
Correct Answer: B
Solution :
(b) \[4.8\times {{10}^{-4}}V\] Here, \[l=6\,cm=6\times {{10}^{-2}}m\] \[B=0.4\,T,\] \[v=2\times {{10}^{-2}}m{{s}^{-1}}\] Voltage developed is \[\varepsilon =Blv\] \[=0.4\times 6\times {{10}^{-2}}\times 2\times {{10}^{-2}}\] \[=4.8\times {{10}^{-4}}V\]You need to login to perform this action.
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