A horizontal frame KLMN moves with a uniform velocity of 30 cm/s into a uniform magnetic field of strength `B=10^(-3)` Tesla acting verticaly downwards. KN = 12 cm and KL = 25 cm and resistance of the frame is `10Omega`. The sides LM and KN enter the field in a direction perpendicular lar to the field boundary. Calculate the current in the metal frame when,
a. LM just enters the field
b. the entire frame inside the field
c. LM just leaves the field through the other side

Explanation

`KL=MN=0.25m" "LM=KN=0.12m" "B=10^(-3)T" "v=0.3m//s`
a. KL and MN are parallel. Hence no induced e.m.f. on these sides. Induced e.m.f. on LM is
`varepsilon_(LM)=Blv=10^(-3)xx0.12xx0.3=36xx10^(-6)V=36muV`
`therefore I=(varepsilon_(LM))/(R)=(36muV)/(10Omega)=3.6muA`
b. When the frame is completely inside, there is no change in flux. i.e., flux through the frame remains constant. Hence there is no induced current.
c. As LM just leaves the field again there is a current, but in the opposite direction.
`therefore I=3.6muA`