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 below. The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit. What is the acceleration of the bar just after the switch S is closed?
2. Relevant equations
Well, the magnetic force of a straight wire is valid here.
F = IlBsinØ
And Newton’s Second Law as well. Fnet = ma.
So combining the two equations, I get IlBsinØ – mg = ma and thus a = (IlBsinØ – mg)/m.
Also, since we have the potential difference of the source, ε = IR is valid as well.
3. The attempt at a solution
The equivalent of the parallel resistors in 5.0 Ω and the total (equivalent of the series) is 30.0 Ω.
ε = IR
I = ε/R
I = (120 V)/(30 Ω)
I = 4.0 A
And the path splits in two before it reaches the bar, so the current that will pass through the bar is 2.0 A.
However, the application of N2L is confusing me. We are not given either the mass of the bar or the acceleration; just the net force of the bar (2.60-N). How do I determine the mass to use in N2L and ultimately obtain the acceleration?