In Fig. shown, the rod has a resistance `R`, the horizontal rails have negligible friction. A battery of emf `E` and negligible internal resistance is connected between points `a and b`. The rod is released from rest.

The velocity of the rod as function of time is

Options

A. (a) `(E)/(Bl)(1 - e^(-t//tau))`
B. (b) `(E)/(Bl)(1 + e^(-t//tau))`
C. ( c) `(3)/(2)(E)/(Bl)(1 - e^(-t//tau))`
D. (d) `(E)/(2Bl)(1 - e^(-t//tau))`

Answer

A

Explanation

(a)Due to applied emf, current will flow as shown. Due to this current and magnetic field `B`, force on the wire will act towards right and rod will start moving towards right. Let at any instant, its velocity, emf `e = Blv` will be induced as shown.

Net emf: `E - Blv, I = (E - Blv)/(R )` ....(i)
`F = IBl = ((E - Blv)/(R ))Bl` ...(ii)
`rarr` `m(dv)/(dt) = (E - Blv)(Bl)/(R )`
`(mR)/(Bl)(dv)/(dt) = E - Blv `rarr` int_(0)^(v)(dv)/(E - Blv) = int_(0)^(t)(Bl)/(mR)dt`
`rarr` `[(1n(E - Blv))/(-Bl)]_(0)^(v) = (Bl)/(mR)t` `rarr` `1n((E - Blv)/(E)) = (-B^(2)l^(2))/(mR)t`
`rarr` `(E - Blv)/(E) = e^(t//tau)` `rarr` `v = (E)/(Bl)[1 - e^(t//tau)]`
Velocity will increase upto `t oo`, and `t = oo`, velocity will be maximum and constant.
Also, when the velocity is constant, net force on the rod will become zero, so putting `F = 0` in equation (ii), we get
`v_(terminal) = (E)/(Bl)`
hen the rod attains terminal velocity, then from equation (i), `I = 0`.