Q152 / 160 · TS EAMCET 2019 (03 May Shift 1) · Physics
A $10 \Omega$ coil of 180 turns and diameter $4 \mathrm{~cm}$ is placed in a uniform magnetic field so that the magnetic flux is maximum through the coil's cross-sectional area. When the field is suddenly removed a charge of $360 \mu \mathrm{C}$ flows through a $618 \Omega$ galvanometer connected to the coil, find the magnetic field.
- A$12 \mathrm{~T}$
- B$6 \mathrm{~T}$
- Correct answerC$1 \mathrm{~T}$
- D$8 \mathrm{~T}$
Solution
Given, resistance of a coil, $R=10 \Omega$ number of turns in the coil, $N=180$ and diameter of the coil $=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$
Magnetic flux associated with the coil is maximum.
Hence, $\theta=0^{\circ}$
$\therefore$ Magnetic flux, $\phi=B A$
When magnetic field is suddenly removed, then some charge flows through the galvanometer.
Given, resistance of a galvanometer, $R_g=618 \Omega$ and charge flowing through the galvanometer, $q=360 \mu \mathrm{C}=360 \times 10^{-6} \mathrm{C}$
As $\quad q=I \cdot \Delta t$
$$
\Rightarrow \quad q=\frac{e}{R_{e q}} \Delta t=\frac{\Delta \phi}{\Delta t} \cdot \frac{\Delta t}{R_{e q}} \quad\left(\because e=N \frac{\Delta \phi}{\Delta t}\right)
$$
So, for $N$ turns, $q=\frac{N}{R_{\text {eq }}} \Delta \phi$
Here, $R_{e q}=R+R_g$
Putting the given values, we get
$$
\begin{aligned}
& 360 \times 10^{-6}=\frac{180}{(618+10)} \cdot(B A-0) \\
& 2 \times 10^{-6} \times 628=B \cdot \pi \times\left(2 \times 10^{-2}\right)^2 \quad\left[\because A=\pi r^2\right] \\
\Rightarrow & B=\frac{2 \times 10^{-6} \times 628}{3.14 \times 4 \times 10^{-4}}=1 \mathrm{~T}
\end{aligned}
$$
Hence, the magnetic field is $1 \mathrm{~T}$.
Hence, the correct answer is option: 'C'.