Magnetic flux through loop inside of solenoid

[as11]

Homework Statement

A solenoid has 1800 turns of wire and is 125 cm long and 10.0 cm in diameter; see Figure P.14 (attached). A circular wire loop of diameter 5.0 cm lies along the axis of the solenoid near the middle of its length as shown.
(a) If the current in the solenoid initially is 4.0 A, find the magnetic flux through the smaller loop.
(b) If the current in the solenoid is switched off and falls to zero in 3.0 s, calculate the average value of the emf induced in the smaller loop.

Homework Equations

1. B(inside solenoid) = \mu₀*n*I, where n=number turns, I= current
2. \Phi_magnetic=\int\stackrel{\rightarrow}{B}*\stackrel{\rightarrow}{dA}

The Attempt at a Solution

\Phi_magnetic=\int\stackrel{\rightarrow}{B}*\stackrel{\rightarrow}{dA} = BA = B*\pi*r² (this did not work, and as I did not get the first part, I had no idea how to even approach the second part...thanks for any advice).

 

[Doc Al]

Why do you say it "didn't work"? Show exactly what you did.

 

[as11]

My apologies.

So B(inside solenoid) = \mu₀*n*I = (1.2566 x 10^-6)*(1800)*(4.0A) = .009048 T

Area loop = \pi*r² = \pi* (6.25 x 10^-4) = .0019635 m²

so, if \Phi= BA = (.009048 T) * (.0019635 m²) = 1.7766 x 10^-5

but this answer was marked incorrect.

 

[Doc Al]

In the equation B = μ*n*I, n is the number of turns per unit length, not just the number of turns.

 

[jdubt]

So to find the flux through the smaller loop do you first find the flux through the solenoid using its area and then somehow use the smaller loop's area to find its flux?

 

[Doc Al]

So to find the flux through the smaller loop do you first find the flux through the solenoid using its area and then somehow use the smaller loop's area to find its flux?

You find the field within the solenoid (which is given by B = μ*n*I), then use it to calculate the flux through the smaller loop's area.

 

[jdubt]

Thanks that worked; I tried that before, but used the diameter instead radius. woops

Part B is giving me trouble as well.

induced emf = -NA * dB/dt

=-1800*(.025)^2 * pi * (-4/3)

 

[jdubt]

Wrong equation, never mind.

induced emf = -L * dI/dt

L= flux / I

 

[as11]

I don't think you have to use an equation; it's asking for the average emf and

emf = -dflux /dt

and dflux= flux2 - flux1 , but flux2 = 0 because the current goes to 0 (stated in problem)

and dt is given in the problem

 

[jdubt]

Your right; you can get the same solution both ways,