Recent content by knowone

I solved these equations: εbattery-εinduced =IR, εinduced = vdB, and m(dv/dt)=IBd Then I= (εbattery-vdB)/R, thus m(dv/dt)=Bd((εbattery-vdB)/R). Using differential equation, v(t)=A(1-e-λt), where A = -εbattery/d2B2 and λ = d2B2/mR Here we added a arbitrary constant R. I'm reluctant to use it...

The power of the battery is εI and it equals Fv=IdBv. Thus εI=idBv and v= ε/dB

Homework Statement I tried to understand the problem b) and c).[/B] Homework Equations Faraday's law: ∇xE = - ∂B/∂t emf ε = Bdv Force : F =ma, Lorenz's force F=q(vxB) ==> ma = IdB Power : power of battery = εI, mechanical power of the wire = Fv The Attempt at a Solution I think I solved...

I'm a SNU student in South Korea. I never had formal physics education before, and now I'm trying to catch up my school curriculum. I hope that I could give tips to other guys later.

Without L, how can I calculate the capacity? I differentiate the electrostatic energy of the conductor to obtain electric force inside the conductor. U= (V^2)C/2 and C changes with the total length of the conductor and the height of the oil. Of course the force do not contain L.

Homework Statement This is the exercise 10.6 from Feynman lectures on Physics 2. Two coaxial pipes of radii a and b(a<b) are lowered vertically into an oil bath. If a voltage V is applied between the pipes, show that the oil rises a height H. Show that H=(V^2)(κ-1)ε_0/[ln(b/a)ρ(b^2-a^2)g] where...