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Q.

In the figure a charge sphere of mass $m$ and charge $q$ starts sliding from rest on a fixed vertical circular track of radius $R$ from the position shown. There exists a uniform and constant horizontal magnetic field of induction $B$ . The maximum force exerted by the track on the sphere is $x m g + q B \sqrt{2 g R}$ . Then the value of $x$ is

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answer is 3.

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Detailed Solution

$θ = \frac{π}{2},$ $f_{m} = q v B,$
And directed radially outward. Since,
$N - m g \sin θ - q B = \frac{m v^{2}}{R}$
Question Image
$\Rightarrow N = \frac{m v^{2}}{R} + m g \sin θ + q v B$
Hence at we get the maximum value of $N$ as
$\Rightarrow N_{\max} = \frac{2 m g R}{R} + m g + q B \sqrt{2 g R}$
$\Rightarrow N_{\max} = 3 m g + q B \sqrt{2 g R}$

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