Electromagnetic Induction And Alternating Current Ques 17

Answer:

Correct Answer: 17.(b)

Solution:

17. Induced constant, $I=\frac{e}{R}$

Here, $e=$ induced emf $=\frac{d \varphi}{d t}$

$$ \begin{aligned} I & =\frac{1}{R}=\frac{d \varphi}{d t} \cdot \frac{1}{R} \\ d \varphi & =I R d t \\ \varphi & =\int I R d t \end{aligned} $$

$\therefore \quad$ Here, $R$ is constant

$\therefore \quad \varphi=R \int I d t$

$$ \begin{aligned} & \int I \cdot d t=\text { Area under } I-t \text { graph } \\ & \quad=\frac{1}{2} \times 10 \times 0.5=2.5 \\ & \therefore \quad \varphi=R \times 2.5=100 \times 2.5=250 \mathrm{Wb.} . \end{aligned} $$