A conducting, sliding crossbar of length L, mass m, and resistance R is mounted

Question

A conducting, sliding crossbar of length L, mass m, and resistance R is mounted on a pair of perfectly conducting, frictionless rails, as shown in the figure below. The crossbar and rails form a complete electric circuit driven by a power supply capable of delivering either a constant emf epsilon 0 or a constant current I0. depending on how the selector switch is set. The entire system is situated in a uniform magnetic field perpendicular to the plane of the circuit, directed into the page. In this first problem, we assume that the power supply is set to deliver a constant current I0. In which direction does the crossbar move, left or right? If the crossbar is initially at rest at time t = 0, derive an expression for the velocity v(t) of the crossbar as a function of time. [Hint: apply Newton's Law using the magnetic force on the crossbar.] Express your result in terms of I0, t, and the parameters of the problem (L, m, R, B). You should find that the velocity increases linearly with time. Since the crossbar is moving, the magnetic flux through the loop of the circuit is changing. By Faraday's Law, there is a corresponding induced emf epsilon ind around the loop. Derive an expression for epsilon ind(t) as a function of time, and carefully determine its direction (clockwise or counterclockwise). [Hint: it is easiest to treat the crossbar as a source of motional emf.] Derive an expression for the time-dependent emf epsilon(t) that the power supply must generate to maintain the constant current I0. [Hint: apply Kirchoff's loop rule to the circuit; the induced emf epsilon ind(t) calculated in part (c) can be represented in the circuit as a separate, localized 'emf source'.] Express your result in terms of I0, t, and the parameters of the problem (L, m, R, B). Write an expression for the time-dependent power P(t) = epsilon(t)I0 delivered by the power supply using your result from part (d). Calculate the total energy delivered by the power supply, and show that it equals the sum of two terms: the total ohmic losses dissipated as heat in the crossbar, and the kinetic energy of the crossbar itself. (See? It's the power supply not the magnetic field - that does the work that goes into the kinetic energy of the crossbar!)

Explanation / Answer

a) towards right

b) F = IoLB

so a = IoLB/m

v =u +at

v(t) = (IoLB/m)*t

c) E = -d(BA)/dt

A = x*L

and dx/dt = v

B is constant

E = -BLv

d) E(t) - Eind = I0*R

E(t) = I0R+ Eind = IoR+(Io(BL)^2 /m)*t =Io[R+((BL)^2 /m)*t]

e)P(t) = Io^2[R+((BL)^2 /m)*t]

f) U(t) = int from 0 to t of P(t) dt

==> U(t) = Io^2 Rt + (IoBLt)^2/2m

==>U(t) = I0^2 Rt + 1/2 mv^2

here IoLBt = mV

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