1. The problem statement, all variables and given/known data Connect a battery to a solenoid A cylindrical solenoid 40 cm long with a radius of 8 mm has 250 tightly-wound turns of wire uniformly distributed along its length (see the figure). Around the middle of the solenoid is a two-turn rectangular loop 3 cm by 2 cm made of resistive wire having a resistance of 105 ohms. One microsecond after connecting the loose wire to the battery to form a series circuit with the battery and a 20 resistor, what is the magnitude of the current in the rectangular loop and its direction (clockwise or counter-clockwise in the diagram)? (The battery has an emf of 9 V.) 2. Relevant equations emf = IR For a solenoid, N = Number of turns I = Current L = Solenoid Length B = μ0*N*I/L n = perpendicular unit vector [itex]\Phi[/itex] = ∫B*nDA emf = d[itex]\Phi[/itex]/dt 3. The attempt at a solution I began by finding the current in the solenoid. 9 V/20 [itex]\Omega[/itex] = .45 Then, I used this current to solve for the magnetic field the solenoid produces. μ0*250*.45/.4 = 3.53*10^-4 Once I found this, I used it to find the magnetic flux. [itex]\Phi[/itex] = 3.53*10^-4*∏*.008^2 = 7.106*10^-8 Then I found d[itex]\Phi[/itex]/dt, which is just the flux over the change in time since starting flux was 0. 7.106*10^-8/(10^-6) = 7.106 * 10^-2 This was the emf, so this *2 is the emf of the rectangular coil (since the rectangular coil has two turns). So emf(coil) = .142122 And the current through the rectangular coil equals this emf divided by its resistance. .142122/105 = .13535 But this wasn't correct.