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

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(C) Given, magnetic field, $B = 0.25 \,T$Mass per unit length, $\frac{m}{l} = 0.5 \,kg \,m^{-1}$Angle of inclination, $	heta = 30^{\circ}$Acceleratio

Vedclass · Accepted Answer

$11.32$

Vedclass · Answer

(C) Given, magnetic field, $B = 0.25 \,T$Mass per unit length, $\frac{m}{l} = 0.5 \,kg \,m^{-1}$Angle of inclination, $	heta = 30^{\circ}$Acceleration due to gravity, $g = 9.8 \,m \,s^{-2}$ (or $10 \,m \,s^{-2}$)The magnetic force $F$ on a current-carrying conductor in a magnetic field is given by $F = BIl$. Since the magnetic field is vertical, the force $F$ acts horizontally.For the rod to remain stationary on the smooth inclined plane, the component of the magnetic force along the plane must balance the component of the gravitational force acting down the plane.Component of gravitational force down the plane = $mg \sin 30^{\circ}$Component of magnetic force up the plane = $F \cos 30^{\circ} = (BIl) \cos 30^{\circ}$Equating the two forces for equilibrium:$BIl \cos 30^{\circ} = mg \sin 30^{\circ}$$BIl \left(\frac{\sqrt{3}}{2}ight) = mg \left(\frac{1}{2}ight)$$BIl \sqrt{3} = mg$$I = \frac{mg}{l \sqrt{3} B} = \left(\frac{m}{l}ight) \frac{g}{\sqrt{3} B}$Substituting the values:$I = 0.5 	imes \frac{9.8}{\sqrt{3} 	imes 0.25} \approx 11.316 \,A \approx 11.32 \,A$Thus, the required current is $11.32 \,A$.

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