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(C) The magnetic Lorentz force acting on the block is given by $F = qvB$,which acts perpendicular to the inclined plane in the upward direction.The bl

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$\frac{m \cot 	heta}{qB}$

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(C) The magnetic Lorentz force acting on the block is given by $F = qvB$,which acts perpendicular to the inclined plane in the upward direction.The block will lose contact with the inclined plane when the upward magnetic force equals the component of the gravitational force perpendicular to the plane:$F = mg \cos 	heta$$qvB = mg \cos 	heta$$v = \frac{mg \cos 	heta}{qB}$The block slides down the incline with an acceleration $a = g \sin 	heta$. Starting from rest,its velocity at time $t$ is given by:$v = at = (g \sin 	heta)t$Equating the two expressions for $v$:$(g \sin 	heta)t = \frac{mg \cos 	heta}{qB}$$t = \frac{mg \cos 	heta}{qB(g \sin 	heta)}$$t = \frac{m \cot 	heta}{qB}$

$A$ block of mass $m$ and charge $q$ is released on a long smooth inclined plane. $A$ magnetic field $B$ is constant,uniform,horizontal,and parallel to the surface as shown. Find the time from the start when the block loses contact with the surface.

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