A) 2.3 m/s
B) 1.2 m/s
C) 0.3 m/s
D) 0.02 m/s
Correct Answer: D
Solution :
[d] The network behaves like a balanced wheatstone bridge. The induced emf developed is given by \[e=vB\ell =v\times 2\times 0.1=0.2v\] ...(i) Now, \[e=IR\] \[e={{10}^{-3}}\times 4=4\times {{10}^{-3}}\,\,\,\,amp\] ...(ii) From (i) and (ii), \[0.2\,v=4\times {{10}^{-3}}\] \[\therefore \,\,\,\,\,v=\frac{4\times {{10}^{-3}}}{0.2}=0.02m/s\]You need to login to perform this action.
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