A metallic rod of mass m and resistance R is sliding over the 2 conducting frictionless rails as shown in Fig. An infinitely long wire carries a current `I_(0)`. The distance of the rails from the wire are b and a respectively.

Find the value of F if the rod slides with constant velocity

Options

A. `(mu_(0)I_(0))/(2 pi R)[E+(mu_(0)I_(0))/(2 pi) v_(0) 1n|(b)/(a)|] 1n ((b)/(a))`
B. `(mu_(0)I_(0))/( pi R)[E-(mu_(0)I_(0))/(2 pi) v_(0) 1n|(b)/(a)|] 1n ((b)/(a))`
C. `(mu_(0)I_(0))/(2 pi R)[E-(mu_(0)I_(0))/(2 pi) v_(0) 1n|(b)/(a)|] 1n ((b)/(a))`
D. None of these

Answer

C

Explanation

(a) Magnetic flux linked with the differential area element shown in fig.

Net flux = `oint d phi = int_(a)^(b) (mu_(0)i)/(2 pi y) xdy = (mu_(0)i)/(2 pi) 1n |(b)/(a)|`
`emf (e) = |-(d phi)/(dt)| = (mu_(0)i)/(2v) v_(0)1n |(b)/(a)|`
`[(dx)/(dt)=v_(0)]`
emf can be calculated using motional emf method, emf developed due to motion of the conductor has an opposite secse to the emf of the cell
therefore, `i=(E_e)/(R)=1/(R){E-(mu_(0)I_(0)v_(0)/(2 pi) 1n |(b)/(a)|}`
(b) `F=int dF=int idvec(L) xx vec(B)`
`=int [1/(R){E-(mu_(0)I_(0)v_(0)/(2 pi) 1n |(b)/(a)|}*dy* (mu_(0)I_(0))/(2 pi y) ]`
`F=(mu_(0)I_(0)/(2 pi) v_(0)1m |(b)/(a)|] 1n ((b)/(a))`.