Consider a straight rod or wire PQ of length l, lying wholly in a plane perpendicular to a uniform magnetic field of induction B , as shown in below figure; \(\vec B\) points into the page.
Suppose an external agent moves the wire to the right with a constant velocity \(\vec v\) perpendicular to its length and to \(\vec B\). The free electrons in the wire experience a Lorentz force \(\vec F\)(= q\(\vec v\)x \(\vec B\)).
A straight wire/rod moving in a uniform magnetic field
According to the right-hand rule for cross products, the Lorentz force on negatively charged electrons is downward. The Lorentz force \(\vec F\) moves the free electrons in the wire from P to Q so that P becomes positive with respect to Q. Thus, there will be a separation of the charges to the two ends of the wire until an electric field builds up to oppose further motion of the charges.
In moving the electrons a distance l along the wire, the work done by the Lorentz force is
W = Fl = (qvB sin θ) l = qvBl
since the angle between \(\vec v\) and \(\vec B\), θ = 90°. Since electrical work done per unit charge is emf, the induced emf in the wire is
e = \(\cfrac Wq\) = vBi
Alternatively, the electric field due to the separation of charges is \(\cfrac{\vec F}q\) = \(\vec v\) x \(\vec B\). Since \(\vec v\) is perpendicular to B, the magnitude of the field = vB.
Electric field = \(\cfrac{p.d. (e) \,between\, P \,and \,Q}{distance \,PQ(l)}\)
Therefore, the p.d. or emf induced in the wire PQ is e = v B l
