A) \[5\mu V\]
B) 5 mV
C) \[50\mu V\]
D) 50 mV
Correct Answer: C
Solution :
(c) \[50\mu V\] The emf developed between the ends of the conductor is \[\varepsilon =\frac{1}{2}\omega B{{l}^{2}}=\frac{1}{2}\times 5\times 0.2\times {{10}^{-4}}\times {{(1)}^{2}}\] \[=5\times {{10}^{-5}}V=50\times {{10}^{-6}}V=50\mu V\]You need to login to perform this action.
You will be redirected in
3 sec