A capacitor of capacitance C = 0.015 F is connected to parallel conducting rail and a conducting rod of mass m = 100 g and length 1m start to fall under gravity in vertical plane. A uniform magnetic field of 2T exist in space directed perpendicular to rod as shown in figure. Find acceleration of rod (m/s2). (use g=10m/s2)
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answer is 0006.25.
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Detailed Solution
mg−BIL=maI=ddt(q)=ddt(CE)=ddt(CBvl) EMF induced in the rod=BV L⇒I=CBLa current in the rod=I mg=BCBLaL+ma∴a=mgm+CB2L2=0.1100.1+0.01541=6.25ms-2