1. The problem statement, all variables and given/known data A 2.60-N metal bar, 1.50 m long and having a resistance of 10.0 Ω, rests horizontally on conducting wires connecting it to the circuit shown in the figure(Figure 1) . The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit. 2. Relevant equations Fb=I LxB Fg=mg I=V/R F=ma 3. The attempt at a solution Resistance of bar and 10 ohm resistor are parallel, so combining them gives a ohm resistor. That is in series with the 25 ohm resistor, so the Req of the circuit is 30 ohms. I=V/R=120/30=4 A Using Kirchoff's laws, found that current passing through bar is 2 A. Since current is moving clockwise, right hand rule says the force due to the magnetic field is moving the bar to the right. The net force is Fb-Fg. Fg is given as 2.6N, Fb is (2)(1.6)x(1.5) Since L and B vectors are perpendicular, take the scalar product. Fb=4.8N To find m, Fg=mg => Fg/g=m => 2.6/9.8=0.27 kg To find acceleration, solve F=ma for a a=F/m So a=(Fb-Fn)/m a=(4.8-2.6)/.027 a=8.15 I got it wrong, and it says that the correct answer is 18.1 m/s^2. Any ideas as to where I went wrong? Thanks.