A) \[2\times {{10}^{4}}V\]
B) \[1.2\times {{10}^{4}}V\]
C) \[2\times {{10}^{-4}}V\]
D) None of these
Correct Answer: A
Solution :
Amount of magnetic flux linked with inductor is \[\phi =Li\] Now, the emf induced in the inductor is given by \[e=-\frac{d\phi }{dt}=-\frac{d}{dt}(Li)\] or \[e=-L\frac{di}{dt}\] or \[\left| e \right|=L\frac{di}{dt}\] Here, induced current \[=\frac{V}{R}\] \[=\frac{10}{5}\] = 2 A Circuit switches off in 1 millisecond or \[dt=1\times {{10}^{-3}}s\] and \[L=10\,H\] \[\therefore \] Induced emf in inductor is \[\left| e \right|=10\times \frac{2}{1\times {{10}^{-3}}}\] \[=2\times {{10}^{4}}V\]
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