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A conducting disc of conductivity $σ$ has radius $a$ and thickness $t$ . If a uniform magnetic field $B$ applied in a direction perpendicular to the plane of the disc, changes with time at the rate of $\frac{d B}{d t} = α$ (Magnetic field is confined to a cylindrical region of radius larger than that of disc, and axis of magnetic field coincides with the axis of the loop)., calculate the power dissipated (in $m W$ ) in the disc due to the induced current.
$[T a k e : α = 2 T / s, a = 1 c m, t = 2 m m a n d σ = \frac{4}{π} \times 10^{8} Ω^{- 1} m^{- 1}]$

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Detailed Solution

Consider an elemental circle of radius r and of thickness dr.

$ϵ = \frac{d}{d t} (π r^{2} B) = π r^{2} α$

$R = \frac{1}{σ} \frac{(2 π r)}{(t d r)}$

For the element the power dissipated inside the path is

$d P = \frac{ϵ^{2}}{R} = \frac{π t σ}{2} a^{2} r^{3} d r$

$\Rightarrow P = \frac{π t σ}{2} α^{2} \int_{0}^{a} r^{3} d r = \frac{π t σ a^{4}}{8} α^{2}$

$\Rightarrow P = \frac{π \times 2 \times 10^{- 3} \times 4 \times 10^{8} \times 10^{- 8}}{8 \times π} \times 4 = 4 m W$

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