A circular conducting coil of radius $1 \mathrm{~m}$ is being heated by the change of magnetic field $\overrightarrow{\mathrm{B}}$ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is $2 \mu \Omega$. The magnetic field is slowly switched off such that its magnitude changes in time as $\mathrm{B}=\frac{4}{\pi} \times 10^{-3} \mathrm{~T}\left(1-\frac{\mathrm{t}}{100}\right)$
The energy dissipated by the coil before the magnetic field is switched off completely is E=________mJ.
$\phi=\vec{B} \cdot \vec{S}$
$\phi=\frac{4}{\pi} \times 10^{-3}\left(1-\frac{\mathrm{t}}{100}\right) \cdot \pi \mathrm{R}^{2}$
$\phi=4 \times 10^{-3} \times(1)^{2}\left(1-\frac{\mathrm{t}}{100}\right)$
$\varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$
$\varepsilon=\frac{-\mathrm{d}}{\mathrm{dt}}\left(4 \times 10^{-3}\left(1-\frac{\mathrm{t}}{100}\right)\right)$
$\varepsilon=4 \times 10^{-3}\left(\frac{1}{100}\right)=4 \times 10^{-5} \mathbf{V}$
When $\mathrm{B}=0$
$1-\frac{t}{100}=0$
$\mathrm{t}=100 \mathrm{sec}$
Heat $=\frac{\varepsilon^{2}}{R} t$
Heat $=\frac{\left(4 \times 10^{-5}\right)^{2}}{2 \times 10^{-6}} \times 100 \mathrm{~J}$
Heat $=\frac{16 \times 10^{-10} \times 100}{2 \times 10^{-6}} \mathrm{~J}$
Heat $=0.08 \mathrm{~J}$
Heat $=80 \mathrm{~mJ}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.