A small block of mass 20 ,g and charge 4 ,mC | vedclass

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(C) Given:Mass of the block, $m = 20 \,g = 0.02 \,kg$Charge on the block, $q = 4 \,mC = 4 	imes 10^{-3} \,C$Magnetic field, $B = 1 \,T$Inclination an

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$5$

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(C) Given:Mass of the block, $m = 20 \,g = 0.02 \,kg$Charge on the block, $q = 4 \,mC = 4 	imes 10^{-3} \,C$Magnetic field, $B = 1 \,T$Inclination angle, $	heta = 45^{\circ}$The magnetic force acting on the moving charged block is given by $F_m = qvB$, which acts perpendicular to the inclined plane (upwards).The block loses contact with the surface when the magnetic force equals the normal component of the gravitational force:$F_m = mg \cos 	heta$$qvB = mg \cos 	heta$$v = \frac{mg \cos 	heta}{qB}$Since the block is released from rest on a smooth inclined plane, its acceleration is $a = g \sin 	heta$. The velocity $v$ at time $t$ is:$v = u + at = 0 + (g \sin 	heta)t = gt \sin 	heta$Equating the two expressions for $v$:$gt \sin 	heta = \frac{mg \cos 	heta}{qB}$$t = \frac{m \cos 	heta}{qB \sin 	heta} = \frac{m \cot 	heta}{qB}$Substituting the values:$t = \frac{0.02 	imes \cot 45^{\circ}}{4 	imes 10^{-3} 	imes 1}$$t = \frac{0.02 	imes 1}{0.004} = 5 \,s$Thus, the block loses contact after $5 \,s$.

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