Well, you could approximate your alternator as a simple inductor, in which case the formula for the field would be B = mu_0 N I / L. mu_0 is the same as before, N is the number of windings, I is the current that flows through, and L is the length of the alternator. However, it is further complicated by the wires being wrapped around an iron core, which will increase the field.
As for the second question, I you need to find the energy density of 87 octane. Wikipedia gives 9.7 kW·h/L, so using $3.40/gallon, you would get $0.09 KwH. However, that assumes you can get 100% of the energy out of your fuel in your engine, which is physically impossible regardless of the engine design. At the temperatures cars run at, the maximum possible efficiency is about 25%, which means in terms of the energy you can actually use, you would be looking at $0.36 KwH. So, the battery at this point is looking about twice as efficient. There are of course many complications. One thing to consider is that car engines are not designed to get as much energy as possible out of the fuel, but instead designed to get lots of horsepower out of the fuel.
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