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A light straight slider 1 m long can slide over two straight long parallel rails, vertically kept. when a uniform magnetic field of 0.6 T is applied perpendicular to the plane of rails, the slider is found moving with a constant velocity.At this instant power dissipated in rails and slider due to their resistance is 1.96 W. If slider has a mass of 0.2kg, then the terminal velocity of the slider is ________.

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Detailed Solution

Let v be the terminal velocity of the slider then, emf developed =e=Blv Power dissipated will be P=i2R=eR2⋅R=eiWhen slider achieve terminal velocity, net downward force on slider is zero.Or magnetic force = gravitational force⇒ Bil=mg⇒ i=mgBl=0.2×9.81.0×0.6=3.27A As power developed P=ie⇒ Emg induced, e=Pi⇒e=1.963.27=0.6V But e=Blv⇒v=eBl=0.60.6×1.0=1ms−1

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A light straight slider 1 m long can slide over two straight long parallel rails, vertically kept. when a uniform magnetic field of 0.6 T is applied perpendicular to the plane of rails, the slider is found moving with a constant velocity.At this instant power dissipated in rails and slider due to their resistance is 1.96 W. If slider has a mass of 0.2kg, then the terminal velocity of the slider is ________.