Induced emf in coil with decreasing current

[kopinator]

A single-turn circular loop of radius R = 0.197 m is coaxial with a long 1740 turn solenoid of radius 0.0410 m and length 0.890 m, as seen in the figure below. (picture in URL)

https://s1.lite.msu.edu/enc/53/3b1bdf0c981a37595901b92ecb54f3656dec7df3ca3110fe53633b89b58fff68fd1816c8653482ff3b80bfab04641cd9644c531bbf37ae59602796c87a446f0a53c5801a82918f44.gif

The variable resistor is changed so that the solenoid current decreases linearly from 6.81 A to 1.59 A in 0.225 s. Calculate the induced emf in the circular loop. (The field just outside the solenoid is small enough to be negligible.)

phi(flux)= ∫B*dA
ε= dphi/dt
A(circle)=∏r^2
B= N*mu_0_*I/L (solenoid)

I'm having troubles finding the flux through the loop. I tried taking the integral from .0410 to .197 m but I don't think that is right. I know once I get my flux integral i can take the flux at both currents, find the difference between the two, and divide by .225 s to find the induced emf. I just don't know what to integrate over.

 

[TSny]

Recall that for a long (ideal) solenoid:
1. B is essentially uniform inside the solenoid
2. B is very weak outside the solenoid so that, to a good approximation, you can assume B = 0 outside the solenoid.

 

[kopinator]

So would i only integrate from 0 to .0410 then?

 

[TSny]

So would i only integrate from 0 to .0410 then?

Yes.

 

[Nickie]

I would suggest using Faraday's Law to calculate the induced emf in the circular loop. Faraday's Law states that the induced emf in a closed loop is equal to the negative rate of change of magnetic flux through the loop. In this case, the magnetic flux through the loop is due to the changing magnetic field created by the solenoid.

To calculate the flux, we can use the formula phi = B*A, where B is the magnetic field and A is the area of the loop. Since the magnetic field just outside the solenoid is negligible, we can use the formula for the magnetic field inside a solenoid B = mu_0*N*I/L, where mu_0 is the permeability of free space, N is the number of turns in the solenoid, I is the current, and L is the length of the solenoid.

Now, to find the area of the loop, we can use the formula A = pi*R^2, where R is the radius of the loop. Plugging in the given values, we get A = pi*(0.197 m)^2 = 0.122 m^2.

So, the magnetic flux through the loop is phi = (mu_0*N*I/L)*A = (4*pi*10^-7 T*m/A)*(1740 turns)*(6.81 A)/(0.890 m)*(0.122 m^2) = 0.00000820 Wb.

To find the induced emf, we can use the formula epsilon = -dphi/dt. Since the current is decreasing linearly, we can use the average current, which is (6.81 A + 1.59 A)/2 = 4.20 A. So, the induced emf is epsilon = -(0.00000820 Wb)/(0.225 s) = -0.0000364 V.

Therefore, the induced emf in the circular loop with decreasing current is approximately 0.0000364 V.