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 ________.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 1.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
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