A metal bar of mass m slides frictionless on two parallel co

A metal bar of mass m slides frictionless on two parallel conducting rails a distance 1 apart (see Fig. below). A resistor R is connected across the rails and a uniform magnetic field B pointing into the page fills the entire region. If the bar moves to the right with velocity v, what is the electromotive force Delta V=-(1/c) in the closed loop. What is the current I in the loop? What is the force on the bar? In what direction? If the bar has velocity v_0 at time t=0, what is its speed at later time t? The power dissipated in the resistor is P = Delta V^2/R. Verify that as t rightarrow infinity all the initial kinetic energy of the bar has been dissipated in the resistor, i.e. (l/2)mV_0^2 = integral P(t)dt (from 0 rightarrow infinity)

Solution

By Faraday\'s law of induction, we know that: EMF = -   / t

a.) In the given situation, the net magnetic flux though the coil = BWL

The rate of change of this flux = d / dt = BL dW/dt = BLv

Therfore, the EMF induced in the loop = BLv

Hence, the current in the loop = EMF / Resistance = BLv / R [Current in the rod will be upwards]

b.) As the bar of length L is carrying a current through it, it will suffer a force given as Current x lenght x magnetic field

That is, F = vB^2L^2 / R. As the current flows upwards in the wire, the direction of the force will be same the that of the vector product of the direction of the current and the magnetic field. That is, towards the left in the figure shown

c.) Due to the action of a force on the rod, it will suffer a deceleration equal to F /m = vB^2L^2 / mR

That is dv/dt = -vB^2L^2 / mR

Rearranging the equation, dv / v = -B^2L^2 dt / mR

Integrating the two sides we get: ln(V/Vo) = -B^2L^2 t/ mR = - kt [Where k = B^2L^2 / mR]

Hence V = Vo e^-kt, where k = B^2L^2 / mR

d.) At any point of time, velocity of the rod = Vo e^-kt

That is the EMF induced = BLVo e^-kt

Therefore Power induced = [BLVo e^-kt]^2 / R

Now dW/dt = [BLVo e^-kt]^2 / R

dW = [BLVo e^-kt]^2 dt / R

Integrating the both sides of the equation above from t = 0 to infinity, we get:

W = [(BLVo)^2 / R] x [-{0 - 1}]/ 2k = [(BLVo)^2 / 2kR] = [B^2 L^2 Vo^2 ] / [2B^2L^2 / m]

W = 0.5*m Vo^2; which indeed is the kinetic energy of the rod at the initial time.