(“A simple magnetic brake”.) In this problem we will examine a braking mechanism

Question

(“A simple magnetic brake”.) In this problem we will examine a braking mechanism based on a combination of induction and the Lorentz force on the conductor with the induced current. In real-life applications, the current is an eddy current similar to the ones running in a copper pipe with a falling magnet. But for simplicity, in this problem we will use a rod being slowed down by a homogeneous magnetic field. Let us consider a conducting rod of mass m sliding without friction over two parallel rails, separated by distance L. The rails are connected via a resistor of resistance R; R is assumed to be much larger than the resistance of either the rod or the rails. The rod is perpendicular to the rails, and the rod is moving in a homogeneous magnetic field B~ , perpendicular to both. The system looks like this:

(a) Express the emf E induced in the rod, as a function of velocity v(t) along the rails. (b) Given the answer to part (a), also express the current I in the circuit comprised of rod + rails + resistor R as a function of v(t)? (c) What is the Lorentz force pulling back the rod as a function of v? Starting from Newtons second law write down the second order differential equation of motion in terms of the distance traveled by the rod which well call x. (So v = dx/dt.) (d) Express the differential equation from part (c) as a first order differential equation of velocity v as the variable. Solve it and obtain v(t). The initial velocity of the rod is v(0) = v0. [Hint: the differential equation is the same as for the RC and RL circuits] (e) How long does it take to reduce the velocity of the rod to 10% of its initial value of v0? (That is, what it the time t10% for which v(t10%) = 0.1v0)?

Explanation / Answer

I am doing first 3 part of the question .Plz post question again to get answer to last 3 parts.

As calculated by Lorenz force ,the emf as a function of velocity is given at

V=B*l*v(t)

where V is emf

B is magnetic field

l is length

and v(t) is velocity

(b) current=emf/resistance=I=BLv(t)/R

(c)the lorenz force pulling back rod is =I*B*L=(BL)^2*v(t)/R

by Newton second law ,mass*acceleration=force=(BL)^2*v(t)/R

m*(d^2(x)/dt^2))=(BL)^2*(dx/dt)/R

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