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## Homework Statement

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.)

## Homework 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

## 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.