What is Motional Electromotive Force e.m.f.? Derive an expression for it.
Definition :
The induced e.m.f. produced due to the motion of a conductor through a uniform magnetic field is called motional e.m.f.
Expression :
Consider a closed circuit or loop PQRS placed in a uniform magnetic field $\vec{B}$ perpendicular to the plane of the paper and directed into the page (shown by $\otimes$). Let $x$ be the length and $l$ be the breadth of the loop PQRS. The arm RS of the loop is free to move on rails, PS and QR without any friction and let it moves with constant velocity $\vec{v}$ towards the left side.
The magnetic flux linked with the loop PQRS is given by
$$\phi_B = BA$$
$$(\because A = lx)$$
$$\phi_B= Blx \tag{i}$$
This magnetic flux changes with time. When arm RS moves to left side with velocity $v$, area enclosed by the loop PQRS decreases.
According to Faraday's law, induced e.m.f. in the circuit is given by
$$ \varepsilon= -\frac{d\phi_B}{dt}$$
$$\varepsilon= -\frac{d}{dt}(Blx)$$
$\varepsilon= -Bl\frac{dx}{dt} \tag{2}$
Since $x$ decreases with time, so
$$-\frac{dx}{dt} = v$$
Where v is speed of the freely moving conductor RS.
Hence,
$\varepsilon = Blv \tag{iii}$
which is the expression for the motional e.m.f. across the conductor RS.
Thus, motional e.m.f. across the conductor is directly proportional to the
[(i)] length ($l$) of the conductor.
[(ii)] speed ($v$) of the conductor.
[(iii)] magnitude of the magnetic field acting perpendicular to the length of the conductor.
If R be the resistance of the closed circuit, then the induced current in the circuit is given by
$I = \frac{\varepsilon}{R} = \frac{Blv}{R} \tag{4}$
Direction of the induced (e.m.f.) current is given with the help of
(i) Fleming's right hand rule
(ii) Thus, induced current in the loop is clockwise.