A) \[0.66\text{ }Wb\text{ }{{s}^{-2}}\]
B) \[1.5\text{ }Wb\text{ }{{s}^{-2}}\]
C) \[-\text{ }0.66\text{ }Wb\text{ }{{s}^{-2}}\]
D) \[-1.5\text{ }Wb\text{ }{{s}^{-2}}\]
E) \[-\,0.33\,Wb{{s}^{-2}}\]
Correct Answer: B
Solution :
From Faradays law, induced emf is \[e=-\frac{d\phi }{dt}\] Given, \[\phi =X{{t}^{2}}\] \[\therefore \] \[e=\frac{-d(X{{t}^{2}})}{dt}=-2t\,X\] Given, \[t=3,\text{ }e=9\text{ }V\] \[\therefore \] \[X=\frac{9}{3\times 2}=1.5\,Wb\,{{s}^{-2}}\]
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