1. The problem statement, all variables and given/known data This has a lot of parts, but I think I got the first few, or at least I'm on the right track. You have an inclined plane (30 degree angle, the ramp part is 3m) with a 1kg metal bar that is 2m long at the top (mu static = mu kinetic = 0.1). The sides of the plane are lined with wires that have no resistance. A light bulb is also in the circuit, which has 0.5 ohms resistance. There is a uniform magnetic field of 0.05T up, which is induced as the bar moves. At the bottom of the ramp, the bar has a velocity of 8m/s. 1. Ignoring the magnetic field, what is the net force on the bar? 2. At the final velocity, what is the induced voltage across the bar? 3. What is the current in the bar? 4. As the bar accelerates, will the magnetic force increase, decrease, or stay the same? 5. what is the magnitude and direction of the magnetic force when the current from #3 is in the bar? 2. Relevant equations Voltage = B*L*velocity V = iR B = (i *mu0) / 2*pi*R 3. The attempt at a solution 1. I used Fnet = Fg - Ff, so (9.81)(1)sin30 - (0.1)(9.81)(1)cos30 = 4.055N Is that right to use the force parallel for Fg? Inclined planes were a long time ago 2. Voltage = B*L*velocity; V = (0.05 T)(3m)(8m/s) = 1.2 V Is this correct? From the later part of this, it seems like the magnetic field would change by the time the bar hits the bottom, meaning it's not 0.05T any more. I don't know how else I can find B though, or if there's another way to find the voltage. 3. i = V/R; i = (1.2V) / (0.5 ohms) = 2.4 Amps 4. Voltage = B*L*velocity, so B = Voltage / (L*velocity), meaning that velocity and the field have an inverse relationship, so as v goes up, B goes down. 5. This is the one that really gets me. I want to use Voltage = B*L*velocity again, but if I convert the current to volts, I'll get the same 0.05T up, which would be fine if the magnetic field stays the same, but it doesn't, does it? I also tried using the third equation, but got a very small number (2.4 * 10^-7) which seemed unreasonably small.