Saved Bookmarks
| 1. |
A conducting bar of 3m length is allowed to fall freely from 80m high tower, keeping it aligned along the East-West direction. Find the emf induced in the rod when it is 20m below the top of tower. g=10 ms^(-2) , B_h=0.7xx10^(-4)T and angle of deep = 60^@ . |
|
Answer» 2.5 mV h=80 m d=20 m `g=10 ms^(-2)` `B_h=0.7xx10^(-4)`T `phi=60^@` Put `v_0=0` in kinetic equation `V^2-v_0^2=2gd` `therefore v^2=2gd` `therefore v^2=2xx10xx20` `therefore v^2=400` `therefore` v=20 m/s `rArr` Kinetic EMF E=Bvl `=B_h COS phi vl` `=0.7xx10^(-4) XX cos60^@xx20xx3` `=0.7xx10^(-4)xx1/2xx20xx3` `=2.1xx10^(-3)` V `therefore` E=2.1 mV |
|