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## Homework Statement

A conducting bar of length L = .218 m and mass M = .08 kg lies across a pair of conducting rails. The contact friction between the bar and the rails is negligible, but there is a resistor at one end with a value R = 20.0 Ohms. Initially the rod is given an initial speed of v0 = 42.0 meters per second. There is a uniform magnetic field perpendicular to the plane containing the rod and rails of magnitude B = 2.2 T.

What is the speed of the rod at time t = 15.303 s?

How far does the rod slide before coming to rest?

## Homework Equations

F=ILB

Ohm’s law: I=V/R

Faraday’s law: V = dΦ/dt = B(dA/dt)

## The Attempt at a Solution

dA = Ldx, giving V=BL(dx/dt)=BLv

F=BLv/R(LB)=(B^2)(L^2)v/R

F=ma=m(dv/dt)

(B^2)(L^2)v/R= m(dv/dt)

If I rearrange this I get:

(B^2)(L^2)/R dt= m/v dv

Taking the integrals of both sides gives:

(B^2)(L^2)t/R (from t=0 to t=t) = m*ln(v) (from v0 to v)

So…

(B^2)(L^2)t/R = m*ln(v) – m*ln(v0)

Solve for v:

v= e^[(B^2)(L^2)t/(Rm)+ln(v0)]

When I put in the numbers I’m not getting the right answer. Is what I did right? Is there something that I’m doing wrong?

Regarding the question on how far the bar goes...I'm guessing that once I get the right equation for velocity, I would set v=0 and solve for t. Then integrate v to find and equation for position and use the t to solve for it. Is this right?

Thanks in advance for any help.