A magnetic field given by $B(t)=\left(0.2 t-0.05 t^2\right) T$ is directed perpendicular to the plane of a…
A magnetic field given by $B(t)=\left(0.2 t-0.05 t^2\right) T$ is directed perpendicular to the plane of a circular coil containing 25 turns of radius $1.8 \mathrm{~cm}$ and whose total resistance is $5 \Omega$. The power dissipation at $3 \mathrm{~s}$ is nearly
$4 \mu \mathrm{W}$
$7 \mu \mathrm{W}$
$2.3 \mu \mathrm{W}$
$1.25 \mu \mathrm{W}$
Solution
Given that, number of turns, $N=25$
Radius of coil, $r=1.8 \mathrm{~cm}=1.8 \times 10^{-2} \mathrm{~m}$
Resistance of coil, $R=5 \Omega$
Magnetic field, $B=\left(0.2 t-0.05 t^2\right) \mathrm{T}$
Now, area of coil,
$\begin{aligned} A & =\pi r^2=3.14 \times\left(1.8 \times 10^{-2}\right)^2 \\ & =1.02 \times 10^{-3} \mathrm{~m}^2\end{aligned}$
Induced emf, $\varepsilon=\frac{-N d}{d t}(B A)=-N A \frac{d B}{d t}$
$\begin{aligned} & =-N A \frac{d}{d t}\left(0.2 t-0.05 t^2\right) \\ \varepsilon & =-N A(0.2-0.1 t)\end{aligned}$
Substituting the values, we get
$\begin{aligned} \varepsilon & =-25 \times 1.02 \times 10^{-3}(0.2-0.1 t) \\ & =-2.55 \times 10^{-2}(0.2-0.1 t)\end{aligned}$
Now, instantaneous power,
$P=\frac{\varepsilon^2}{R}=\frac{\left[-2.55 \times 10^{-2}(0.2-0.1 t)\right]^2}{5}$
At, $t=3 \mathrm{~s}$, power dissipated is
$P=\frac{\left(2.55 \times 10^{-2}\right)^2(0.2-0.1 \times 3)^2}{5}$
$=1.30 \times 10^{-6} \mathrm{~W}=1.30 \mu \mathrm{W}$
It is close to option (d).