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.)
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
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.