A `0.5m` long metal rod `PQ` completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of flux density `0.15 T`. If the resistance of the total circuit is `3Omega` calculate the force needed to move the rod in the direction as indicated with a constant speed of `2 ms^(-1)`

Explanation

Emf induced across a rod of length l, moving with velocity v in an external magnetic field B is
`epsilon = Blv`
Induced current will be
`I=(epsilon)/(R)`
`"where R is the resistance of the circuit" = (Blv)/(R)`
Magnitude of force acting on a current carrying conductor of length l, moving in magnetic field is
`F=IlBsintheta`
Here, `theta=90^(@)`
`implies F=IlB=(Blv)/(R).lB=(B^(2)l^(2)v)/(R)`
Given, B=0.15 T, l=0.5 m, `v=2 m // s` , `R=3 Omega`
`implies F=((0.15)^(2) xx (0.5)^(2) xx 2)/(3)=3.75 xx 10^(-3) V`