A circular coil of radius $10 \mathrm{~cm}$ and resistance of $2 \Omega$ is placed with its plane…
A circular coil of radius $10 \mathrm{~cm}$ and resistance of $2 \Omega$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180^{\circ}$ in $0.25 \mathrm{~s}$. If the magnitude of the induced emf is $3.8 \times 10^{-3} \mathrm{~V}$, then the number of turns of the coil is
(Horizontal component of earth's magnetic field at the place is $3 \times 10^{-5} \mathrm{~T}$ )
504 turns
458 turns
302 turns
608 turns
Solution
Radius of coil, $\mathrm{r}=10 \mathrm{~cm}=0.1 \mathrm{~m}$
Resistance of coil, $R=2 \Omega$
Angle of rotation, $\theta=180^{\circ}$
Time of rotation, $\mathrm{T}=0.25 \mathrm{sec}$
Inducided emf, $\mathrm{E}=3.8 \times 10^{-3} \mathrm{~V}$
Horizontal Magnetic field, $\mathrm{B}_{\mathrm{H}}=3 \times 10^{-5} \mathrm{~T}$
Induced emf is given as,
$\begin{aligned} & E=-\frac{N\left(\phi_f-\phi_i\right)}{t} \\ & =\frac{N\left(\phi_i-\phi_f\right)}{t} \\ & =\frac{N\left(B A \cos \theta^{\prime}-B A \cos \theta\right)}{t}\end{aligned}$
$\theta^{\prime}=$ Initial angle $=0^{\circ}$
Substitute necessary values,
$3.8 \times 10^{-3}=\frac{\mathrm{N}\left[3 \times 10^{-5} \times \pi \times(0.1)^2+3 \times 10^{-5} \times \pi(0.1)^2\right]}{0.25}$
on rearranging \& solving,
Number of turns $\mathrm{N}=504$