A) 1 V
B) 11 V
C) 9 V
D) 10 V
Correct Answer: C
Solution :
Let the net e.m.f.in the circuit be\[{{\varepsilon }_{net}}\] Now,\[{{\varepsilon }_{net}}={{\varepsilon }_{ind}}+{{\varepsilon }_{b}}\] Where \[{{\varepsilon }_{ind}}\]is the induced e.m. fand \[{{\varepsilon }_{b}},\]the e. m. f. of the battery. Applying Faraday's law \[{{\varepsilon }_{ind}}=-\frac{d\phi }{dt}\] If we make anticlockwise circumvention in the loop then the area vector is vertically upward and the magnetic field is vertically downward. Now, \[\phi =\vec{B}.\vec{A}=-BA\] \[\therefore \] \[\phi =-(0.01-2t)\frac{1}{2}\Rightarrow \frac{d\phi }{dt}=1\,V\] So, \[{{\varepsilon }_{ind}}=-\frac{d\phi }{dt}=-1\,V\] \[\therefore \]\[{{\varepsilon }_{net}}={{\varepsilon }_{ind}}+{{\varepsilon }_{b}}=-1+10=9\,V\] Hence, the correction option is [c].
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