Question
The figure shows a conducting loop consisting of half circle of area $$A = 0.06\,{m^2}$$ and three straight segments. The half circle lies in a uniform changing magnetic field $$B = 4{r^2} + 2t + 5$$ ($$SI$$ unit), where $$t$$ is the time in second. An ideal battery $$E = 2\,V$$ is connected as shown and the total resistance of the wire is $$2\Omega .$$ The net current in the loop is at $$t{\text{ }} = {\text{ }}5{\text{ second}}$$ is:
The figure shows a conducting loop consisting of half circle of area $$A = 0.06\,{m^2}$$ and three straight segments. The half circle lies in a uniform changing magnetic field $$B = 4{r^2} + 2t + 5$$ ($$SI$$ unit), where $$t$$ is the time in second. An ideal battery $$E = 2\,V$$ is connected as shown and the total resistance of the wire is $$2\Omega .$$ The net current in the loop is at $$t{\text{ }} = {\text{ }}5{\text{ second}}$$ is:
A.
$$1\,A$$
B.
$$1.5\,A$$
C.
$$0.26\,A$$
D.
$$0.10\,A$$
Answer :
$$0.26\,A$$
Solution :
$$\eqalign{ & {e_{\operatorname{ind} }} = \frac{{d\phi }}{{dt}} = \frac{d}{{dt}}\left( {BA} \right) = A\frac{{dB}}{{dt}} \cr & = A\frac{d}{{dt}}\left( {4{t^2} + 2t + 5} \right) = A\left( {8t + 2} \right) = 0.06\left( {8t + 2} \right) \cr & = 2.52\,{\text{at}}\,t = 5 \cr & i = \frac{{2 - {e_{{\text{ind}}}}}}{2} = 0.26\,A \cr} $$
$$\eqalign{ & {e_{\operatorname{ind} }} = \frac{{d\phi }}{{dt}} = \frac{d}{{dt}}\left( {BA} \right) = A\frac{{dB}}{{dt}} \cr & = A\frac{d}{{dt}}\left( {4{t^2} + 2t + 5} \right) = A\left( {8t + 2} \right) = 0.06\left( {8t + 2} \right) \cr & = 2.52\,{\text{at}}\,t = 5 \cr & i = \frac{{2 - {e_{{\text{ind}}}}}}{2} = 0.26\,A \cr} $$
