A long straight wire carries a current `I_(0)`. At distances `a` and `b` from it there are two other wires, parallel to the former one, which are interconnected by a resistance `R` (Fig). A connector slides without friction along the wires with a constant velocity `v`. Assuming the resistances of the wires, the conductor, the sliding contacts, and the self-inductance of the frame to be negligable, find:
(a) the magnitude and the direction of the current induced in the connector,
(b) the force required to maintain the connector's velocity constant.

Explanation

As `vec(B)`, due to the straight current carrying wire, varies along the rod (connector) and entres lineraly so, to make the calculations simple, `vec(B)` is made constant by taking its average value in the range `[a,b]`
`lt B gt = (int_(a)^(b) B dr)/(int_(a)^(b) dr) = (int_(a)^(b) (mu_(0))/(2pi) (i_(0))/(r) dr)/(int_(a)^(b) dr)`
or, `lt B gt = (mu_(0))/(2pi) (i_(0))/((b - a)) In (b)/(a)`

(a) The flux of `vec(B)` changes through the loop due to the movement of the connector. According to Lenz's law, the current in the loop will be anticlockwise. The magnitude of motional e.m.f.
`xi_(m) = v lt B gt (b - a)`
`= (mu_(0))/(2pi) (i_(0))/((b - a)) In (b)/(a) (b - a) (dx)/(dt) = (mu_(0))/(2pi) i_(0) In (b)/(a) v`
So, induced current
`i_(i n) = (xi_(i n))/(R) = (mu_(0))/(2pi) (i_(0))/(R) In (b)/(a)`
(b) The force required to maintain the constant velocity of the connector must be the magnitude equal to that of Ampere's acting on the connecor, but in opposite direction.
So, `F_(ext) = i_(i n) l lt B gt = ((mu_(0))/(2pi) (i_(0))/(R) v In (b)/(a)) (b - a) ((mu_(0))/(2pi (i_(0))/((b - a))) In (b)/(a))`
`= (v)/(R) ((mu_(0))/(2pi) i_(0) In (b)/(a))^(2)`, and will be directed as shwon in the (fig).