Ranking induced emfs and currents {Faraday's Law}

[skwz]

Hey everyone, I'm having trouble with b and c and any help would be appreciated! I attached two pictures on the bottom (one of the entire problem and one of just the diagram)

Homework Statement

Five loops are formed of copper wire of the same gauge (cross-sectional area). Loops 1-4 are identical; loop 5 has the same height as the others but is longer. At the instant shown, all the loops are moving at the same speed in the directions indicated

There is a uniform magnetic field pointing out of the page in region 1; in region 2 there is no magnetic field. Ignore any interactions between the loops

a.) For any loop that has and induced current, indicate the direction of that current

b.) Rank the magnitudes of the emfs around the loops. Explain your reasoning

c.) Rank the magnitudes of the currents in the loops. Explain your reasoning

http://img412.imageshack.us/img412/4615/diagramfaraday1yp1.th.jpg

Homework Equations

$$\xi = - \frac{d \Phi_B}{dt} = - \frac{d}{dt} (B \ell x) = -B \ell \frac{dx}{dt} = -B \ell v$$

$$I = \frac{\xi}{R}$$

$$R = \rho \frac{\ell}{A}$$

The Attempt at a Solution

a.) I got:
$$I_1$$ is counterclockwise
$$I_2$$ is zero
$$I_3$$ is zero
$$I_4$$ is clockwise
$$I_5$$ is clockwise

b.) $$\xi_1 = \xi_4 = \xi_5 = \xi_3 = \xi_2$$

Since magnetic flux varies with respect to time in loops 1,5, and 4, I used $$\xi = -B \ell v$$ to get the magnitudes of these loops (Loops 2 and 3 have zero emf since the magnetic flux doesn't change with respect to time). Since B is constant and all the loops are moving at the same velocity, then $$\ell$$ determines the magnitude. Now is $$\ell$$ the length of the whole loop or the length across the emf (potential difference), which in case $$\ell$$ would only by the height of loops 1, 4, and 5. If the latter is true, then I believe I have the correct answer.

c.) $$I_5 = I_4 = I_1 > I_2 = I_3$$

There is no induced current in loop 2 and 3. To calculate the magnitudes of the remaining loops I relied on $$R = \rho \frac{\ell}{A}$$.
Since we given that all the loops are made of copper and they all have the same cross sectional area, then A and $$\rho$$ remain constant. Now here is where I'm having trouble again, is $$\ell$$ referring to the length of the entire loop or the distance across the emf, which would again be the height. If it's the latter, then resistance is the same for all the loops, meaning that $$I \propto \xi$$ and thus $$I_5 = I_4 = I_1$$.

 

[TSny]

Your answers for part (a) are correct.

b.) $$\xi_1 = \xi_4 = \xi_5 = \xi_3 = \xi_2$$

I believe you have a typo here. The third = sign should be >. You stated correctly that $\xi_3 = \xi_2 = 0$.

Now is $$\ell$$ the length of the whole loop or the length across the emf (potential difference), which in case $$\ell$$ would only by the height of loops 1, 4, and 5.

Yes, that's right.

c.) $$I_5 = I_4 = I_1 > I_2 = I_3$$

There is no induced current in loop 2 and 3. To calculate the magnitudes of the remaining loops I relied on $$R = \rho \frac{\ell}{A}$$.
Since we given that all the loops are made of copper and they all have the same cross sectional area, then A and $$\rho$$ remain constant. Now here is where I'm having trouble again, is $$\ell$$ referring to the length of the entire loop or the distance across the emf, which would again be the height. If it's the latter, then resistance is the same for all the loops, meaning that $$I \propto \xi$$ and thus $$I_5 = I_4 = I_1$$.

The current in a loop depends on the total resistance of the loop. So, $\ell$ in the R formula should be the total length of the loop. The current in #5 will not be the same as in #1 and #4.