A coil is placed in a time varying magnetic field. The power dissipated due to current induced in the coil…
A coil is placed in a time varying magnetic field. The power dissipated due to current induced in the coil is $P_1$. If the number of turns is doubled and radius of the wire is halved, the power dissipated is $P_2$. Then $P_1: P_2$ is
1 : 4
3 : 2
2 : 1
4 : 1
Solution
According to the question, radius of wire become $\frac{r}{2}$, so its length will be $4 l$ and its resistance will become $16 R$. The number of turns is doubled in a coil, so its radius should be doubled to accomodate the length of wire. The area of coil will become 4 times.
Now, current induced in the coil is $P$,
$
\therefore \quad P_1=\frac{V_1^2}{R}
$
From Eqs. (i) and (ii), we get
$
\begin{gathered}
P_2=\frac{\left(8 V_1\right)^2}{16 R} \Rightarrow P_2=\frac{64 V_1^2}{16 R} \\
P_2=\frac{4 V_1^2}{R}
\end{gathered}
$
Then the ratio, $P_1: P_2=\frac{V_1^2}{R}: \frac{4 V_1^2}{R}$ or $P_1: P_2=1: 4$
