Question

Two long parallel horizontal rails, a distance d apart and each having a resistance $\lambda $ per unit length, are joined at one end by a resistance R. A perfectly conducting rod MN of mass m is free to slide along the rails without friction (see figure). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that, as the rod moves, a constant current i flows through R. Find the velocity of the rod and the applied force F as functions of the distance x of the rod from R.v

Answer 1

Total resistance of the circuit as function of distance x from resistance R is
$ \, \, \, \, \, \, \, R_{net} = R + 2 \lambda x$
Let v be velocity of rod at this instant, then motional emf induced across the rod,
$ \, \, \, \, \, \, \, $ e = Bvd
$\therefore \, \, Current i = \frac{e}{R_{net}} = \frac{Bvd}{ R + 2 \lambda x}$
$ \therefore \, \, \, \, v = \frac{(R + 2 \lambda x ) i}{Bd} $
Net force on the rod,
$ \, \, \, \, \, \, F_{net} = m \frac{dv}{dt}$
$ \, \, \, \, \, \, \, \, \, = \frac{2 \lambda im}{Bd} (R + 2 \lambda) \cdot \frac{dx}{dt}$
$but \, \, \, \, \, \, \, \, \frac{dx}{dt} = v = \frac{(R + 2 \lambda x)i}{Bd}$
$F_{net} = \frac{2 \lambda i^2 m}{B^2 d^2}(R + 2 \lambda x)^2$
This net force is equal to $F - F_m$ where $F_m$ = idB
$ \therefore \, \, F= F_{net} + F_m = \frac{2 \lambda i^2 m}{B^2 d^2} (R + 2 \lambda x)^2 + idB$