Induced current in a loop of wire when straight wire is cut

[astrocytosis]

Homework Statement

A square loop, side a, resistance R, lies a distance s from an infinite straight wire that carries current I (pointing to the right). Now someone cuts the wire, so I drops to zero. In what direction does the induced current in the square loop flow, and what total charge passes through a given point in the loop during the time the current flows?

Homework Equations

$$B_{wire} = \frac{\mu_{0}I}{2\pi r}$$
$$\oint\vec{E} \cdot d\vec{l} = -\frac{\partial}{\partial t} \int \vec{B} \cdot d\vec{a}$$

The Attempt at a Solution

When the wire is cut, the flux is going to decrease, so there must be a current flowing counterclockwise in the loop to oppose the change in flux.

I think I have to use Faraday's law to find out something about the electric field induced in order to get the total charge passing through a given point, but I'm having trouble getting started. The electric field must be circulating around the wire, so it has a curl, but I'm not sure how to find it. Differentiating B with respect to t doesn't make much sense to me either. In general I am confused about the form of the induced electric field.

 

[RedDelicious]

Counterclockwise sounds good

Faraday's law also sounds good. What is Faraday's Law? How is it related to flux? Can you find the flux? In this problem you want to find charge. Can you think of how to relate Faraday's law to charge. Hint: think of the loop.

 

[astrocytosis]

Faraday's law states that EMF is the negative of the time derivative of the magnetic flux. I can find the flux for the case where there is current flowing through the wire:

$$\Phi = \frac{ \mu_{0} I}{2\pi} \int_{s}^{s+a} \frac{a dr}{r} = \frac{ \mu_{0} I}{2\pi} a ln(\frac{s+a}{s})$$

But it's taking the time derivative of this that confuses me, since the cut occurs instantaneously.

The EMF is related to the current, which is related to the charge, by I = EMF/R → dQ/dt = EMF/R

So I could write

$$\int dQ = \frac{1}{R} \int EMF dt$$

and since

$$ EMF = -\frac{d\Phi}{dt},$$
then
$$\int EMF dt = -\Phi$$

which means I can plug into to my equation for charge to get

$$Q = \frac{1}{R} (-\Phi)$$

So the charge would just be the negative flux over the resistance. Is that right?

 

[rude man]

Very good!
But you didn't really specify the direction of loop current flow. It's either clockwise or counterclockwise depending on whether the loop sits above or below the wire. Safe way is to go with Lenz's law.