A uniform disc of radius r and mass m is charged uniformly with the charge q. This disc is placed flat on a rough horizontal surface having coefficient of friction $μ .$ A uniform magnetic field is present in a circular region (a > r) but varying as $k t^{3}$ as shown in figure. Find the time in second after which the disc begins to rotate. (Given r = 1m, m = 18 kg, q = 1C, $μ = 0.1,$ K = 4, g = 10 $m / s^{2}$ )

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Detailed Solution

$E = \frac{x}{2} \frac{d B}{d t}$
$E = \frac{3 K x t^{2}}{2}$
$d t = \frac{3 K x t^{2}}{2} \times \frac{2 π x d x}{π r^{2}} q . x$
Question Image
$τ = \frac{3 K t^{2} q}{r^{2}} \int_{0}^{r} x^{3} d x$
$τ = \frac{3 K q . t^{2}}{4}$ …..(i)
torque due to friction force
$d τ = μ d m g x$
$τ = 2 μ g \frac{q m}{r^{2}} \int_{0}^{r} x^{2} d x = \frac{2}{3} μ m g r$ …..(ii)
$\frac{3 K q . t^{2} r^{2}}{4} = \frac{2}{3} μ m g r$
$t = \sqrt{\frac{8 μ m g}{9 K q r} = 2}$ seconds