Two long parallel conducting horizontal rails are connected by a conducting wire at one end. A uniform magnetic field B (directed vertically downwards) exists in the region of space.

A light uniform ring of diameter d which is practically equal to separation between the rails is placed over the rails as shown in Fig. If resistance of ring be `(lambda)` per unit length
The force required to pull the ring with uniform velocity v is

Options

A. `(4 B^(2)vd)/(3 pi lambda)`
B. `(3 B^(2)vd)/(4 pi lambda)`
C. `(2 B^(2)vd)/(3 pi lambda)`
D. `(4 B^(2)vd)/(pi lambda)`

Answer

D

Explanation

When ring moves to the right, emf is induced in each of the two semi-circles. During an elemental time interval dt, displacment of the ring is v dt.
Consider tow semi-circular separately.
During this interval, left semi-circle cuts flux area shown as shaded part in fig. This area is d (v dt). Therefore, flux cut by semi-circle during this interval is,

`d phi = Bd (vdt)`
Hence, emf induced in it is `e=(d phi)/(dt) = Bvd`
This emf tries to force an anticlockwise current in the circuit as shown in fig.
Similarly, emf induced in right semi-circle is also equal to `e=Bvd` and it also tries to force current in the same direction as shown in fig.

Hence , these tow semi-circles are two identicla electrical sources connected in parallel with each other. EMF of each source is `e=Bvd` and internal resistance is, `r=(pi d//2)lambda`.
`:.` Equivalent internal resistance of parallel combination of two source is
`r/2 = 1/4 pi d lambda`
Since, rails have negligible resistance, therefore, equivalent resistance of the circuit is,
`R=r/2 = 1/4 pi lambda d`
Hence induced current through rails is
`l=e/R`
But current through each semi-circle is equal to `I//2`.
Force required to maintain velocity of ring at constant = Retarding force acting on ring due to induced current = 2 X Retarding force on each semi-circle. ltbr. `2 xx B 1/2 d = Bld = (4 B^(2)vd)/(pi lambda)`.