Q124 / 160 · TS EAMCET 2017 · Physics
A coil having $n$ turns and resistance $R \Omega$ is connected with a galvanometer of resistance $4 R \Omega$. This combination is moved in time $t$ seconds from a magnetic flwx $\phi_1$ Weber to $\phi_2$ Weber. The induced current in the circuit is
- A$\frac{\phi_2-\phi_1}{5 \operatorname{Rnt}}$
- Correct answerB$-\frac{n\left(\phi_2-\phi_1\right)}{5 \text { Rt }}$
- C$-\frac{\left(\phi_2-\phi_1\right)}{\text { Rint }}$
- D$-\frac{n\left(\phi_2-\phi_1\right)}{R t}$
Solution
Given,
Number of turns in coil $=n$
We krsow that,
$$
\begin{aligned}
& \text { Current (i) }=\frac{e}{R^{\prime}}=\frac{-\Delta \phi}{R^{\prime} \Delta t} \\
& {\left[\because a=\frac{-\Delta \phi}{\Delta t}\right]} \\
& R=R+4 R \\
& i=-\frac{\left(\phi_2-\phi_1\right)}{(R+4 R) \Delta t} \\
& i=\frac{\left(\phi_2-\phi_1\right)}{5 R(\Delta t)} \\
&
\end{aligned}
$$
Here, coil have n number of turns.
Hence, $\quad i=-\frac{n\left(\phi_2-\phi_1\right)}{5 R(\Delta t)}$
Induced current, $i=-\frac{n\left(\phi_2-\phi_1\right)}{5 R t}$
Hence, the correct answer is option: 'B'.