Figure shows a conducting rod PQ in contact with metal rails RP and SQ which are 0.257 apart in i uniform magnetic field of flux density 0.4T acting perpendicular to the plane of the paper. Ends R and S are connected through a 5 `Omega` resistance. What is the emf when the rod moves to the right with a velocity of 5 `ms^(-1)`? What is the magnitude and direction of the current through the 5 `Omega` resistance? If the rod PQ moves to the left with the same speed, what will be the new current and its direction?

Explanation

`|e| = Blv = 0.4 xx 0.25 xx 5 = 0.5V`
Current , `I = (|e|)/( R) = (0.5V)/(5Omega) = 0.1A`
As the rod .PQ. moves to right as shown, the free electrons in it experience a Lorentz force. According to F.L.H., the force is towards the end .Q. of rod. `therefore ` They move from P to Q, hence the end of the rod P becomes deficient of electrons
`rArr V_P gt V_Q`
Applying Fleming.s right hand rule, the current in the rod shall flow from Q to P.
(b): If the rod PQ moves to the left with the same speed, then the current of 0.1 A will flow in the rod PQ from P to Q