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A uniform disc of radius $r$ and mass $m$ is charged uniformly with 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 $B = k t^{3}$ as shown in fig. Then the time (in second) after which the disc begins to rotate is_________
$(G i v e n r = 1 m, m = 18 k g, q = (1 / 3) C, μ = 0.1, k = 4 (S . I . u n i t s), g = 10 m / s^{2})$

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answer is 3.46.

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Detailed Solution

Included $E = \frac{x}{2} \frac{d B}{d t}$ $E - f i e l d E = \frac{3 k x t^{2}}{2}$

For small elemental part $d τ = \frac{3 k x t^{2}}{2} q \times \frac{2 π x d x}{π r^{2}} x$

$d τ = \frac{3 k t^{2} q}{r^{2}} \int_{0}^{r} x^{3} d x = \frac{3 k q t^{2} r^{2}}{4}$

Torque due to friction force $τ = \int_{0}^{r} d τ = \frac{2}{3} μ m g r$

$\frac{3 k q t^{2} r^{2}}{4} = \frac{2}{3} μ m g r \Rightarrow t = \sqrt{12} \sec$

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