emf = dΦ/dt = (B*A)*d/dt = B(dA/dt), dA/dt= L*d/dt(vt) = L*v, emf = B*L*v per coil
Since there are 25 loops the total emf= 25(vBL) This is where I'm am stuck. Would I assume that B is 24 uT, the velocity as 3m/s , and the length as 1mm? If so I would get ∆V as 1.8*10^-6.
V=I*R
6v=I*(0.6+0.9)ohms
I=4amp
B=100*(uo)(2N)(I)/L * 1/2 I think since the wire is double wrapped, we multiply the equation by 2, but since we are looking for the magnetic field at the end of the wire we also have to multiply the equation by 1/2
I=4A, uo= 4pi*10^-7
2N/L turns per unit...