A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards left with a velocity of 10 m .`s^(-1)`, calculate the emf induced in the arm. Given the resistance of the arm to be `5 Omega`, (assuming that the other arms are of negligible resistance) find the value of the current in the arm.

Explanation

In unit time, the arm MN describes an area lv. So, the rate of change of magnetic flux linked with MN is
`(dphi)/(dt)=(d)/(dt)(Bxx"area")=B(d)/(dt)("area")=Blv`
`therefore` The magnitude of emf induced in the arm (omitting the negative sign) is
`|e|=(dphi)/(dt)=Blv=0.5xx(20xx10^(-2))xx10=1V`
`therefore` Induced current in the arm MN,
`i=(1)/(5)=0.2A`