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)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 0006.25.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

Watch 3-min video & get full concept clarity