Question

Following figure shows a metal rod PQ resting on the smooth ralls AH and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod 15 cm, B 0.50 T, resistance of the closed loop containing the rod `9.0 mOmega`. Assume the field to be uniform.
What is the retarding force on the rod when K is closed?

Answer 1

Given, length of the rod `l=15cm=15xx10^(-2)m` Magnetic field `B=0.50T`
Resistance of the closed - loop containing the rod, `R=9mOmega=9xx10^(-3)Omega`
Velocity of rod `V=12" cm/s "12 xx 106(-2) m//s.`
When the key Kis closed, current flow in the loop and the current carrying wise experience a retarding force in the magnetic field which is given by `"Force"=BIl=B.e/R.l =(0.5xx9xx10^(-3)xx15xx10^(-12))/(9xx10^(-3))=7.5xx10^(-2)N`