A magnetic field `B = B_(0) sin (omega t) hat k` covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d, Fig. The wires are in the x-y plane. The wire AB (of length d) has resistance R and parallel wires have negligible resistance. If AB is moving with velocity `upsilon`, what is the current in the circuit ? What is the force needed to keep the wire moving at constant velocity ?

Explanation

In Fig. we have shown a wire AB of length l -d, resistance R moving with a velocity `upsilon` along positive direction of X-axis over two parallel wires OX and CD at y = 0 and y = d respectively.
The magnetic field covering the region is `vec B = B_(0) sin (omega t) hat k`.
At t = 0, AB is at x = 0 and velocity = `upsilon hat i` , At t = t, AB is at x = `upsilon` t
Motional emf in AB = `(B_(0) sin omega t) d.v. (-hat j)`.
Its direction, as per Fleming's right hand rule is along negative y-axis.
e.m.f. due to change in magneticc field `= - B_(0) omega cos omega t (x) (d)`
It is also along negative y-axis.
`:.` total e.m.f. induced `=B_(0) sin omega t d.v. = B_(0) omega cos (omega t) x.d = - B_(0) d[upsilon sin omega t + omega x cos omega t]`
Current induced, `I = (B_(0) d)/(R ) [upsilon sin omega t + omega x cos omega t]`
The direction of current is clockwise, i.e., ABOC (absorbing negaitve sign.)
`:.` Force needed along `hat i` , F = B I L
`F = B_(0) sin omega t. (B_(0) d)/(R) [ upsilon sin omega t + omega x cos omega t] d`
`F = (B_(0)^(2) d^(2))/(R ) [upsilon sin omega t + omega x cos omega t] sin omega t`