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Magnetic Force, Resistance, Faraday's Law |
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| Mar15-07, 06:01 PM | #1 |
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Magnetic Force, Resistance, Faraday's Law
1. The problem statement, all variables and given/known data
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? 2. Relevant equations F=ILB Ohm’s law: I=V/R Faraday’s law: V = dΦ/dt = B(dA/dt) 3. 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. |
| Mar16-07, 01:20 AM | #2 |
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good problem, i think. not sure why you haven't some help. Likely tomorrow am.
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