A metallic rod of mass per unit length 0.5; k | vedclass

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(C) The force due to gravity acting down the incline is $F_g = mg \sin 30^{\circ}$.The magnetic force acting on the rod is $F_m = IlB$,which acts hori

Vedclass · Accepted Answer

$11.32$

Vedclass · Answer

(C) The force due to gravity acting down the incline is $F_g = mg \sin 30^{\circ}$.The magnetic force acting on the rod is $F_m = IlB$,which acts horizontally. The component of this magnetic force acting up the incline is $F_{m, 	ext{up}} = IlB \cos 30^{\circ}$.For the rod to remain stationary,the forces along the incline must balance:$mg \sin 30^{\circ} = IlB \cos 30^{\circ}$Rearranging for current $I$:$I = \frac{mg}{lB} 	an 30^{\circ}$Given mass per unit length $\frac{m}{l} = 0.5\; kg\; m^{-1}$,$B = 0.25\; T$,and $g = 9.8\; m/s^2$:$I = \frac{0.5 	imes 9.8}{0.25} 	imes 	an 30^{\circ}$$I = 2 	imes 9.8 	imes \frac{1}{\sqrt{3}}$$I = \frac{19.6}{1.732} \approx 11.32\; A$

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