# Finding induced current in a loop at an instant [closed]

I'm working on this problem; however, I cannot seem to get anywhere.

Given information:

The rectangular loop in the figure has 2.1x10^-2 ohm resistance. What is the induced current in the loop at this instant?

It asks for the answer of the current of the loop in terms of I of the rod.

I attempted the problem:

I_loop = emf/R = vLBA / R

emf = d(magnetic flux)/dt = d(B * A)/dt

B = (mu_0 I)/(2 pi d)

From here I'm not exactly sure what to do. Any ideas?

-

## closed as too localized by Qmechanic♦Feb 15 '13 at 15:34

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Your way to solve this problem is correct. The basic idea is calculate the voltage induced by cutting the magnetic field. The induce potential is

$U = \frac{d\Phi}{dt} = \frac{d}{dt}\iint_{A}\vec{B}d\vec{A}$

since the magnetic field $\vec{B}$ is a function of the distance from wire to any point in space $r$, so you cannot easily say $\Phi = B\cdot A$. Instead, you should integral $B$ over the whole area of that rectangular loop.

So, here, we know the magnetic field is a function of $r$, and the direction of magnetic field in the plane of this loop is parallel to the direction of normal vector of the surface, so we can not $B$ as $B(r)$. Therefore,

$U = \frac{d}{dt}a \int_{r_1}^{r_2}B(r)dr$

where $a$ is the height of the rectangular loop ($a = 4cm$). Then you can solve this problem.

By the way, there is one easy method to solve this:

Since the direction of magnet field is always perpendicular to the plain that this loop is in, so when you move the loop, only two edges of the loop that are parallel to the current is cutting the magnetic field, therefore, only this edge will have induced potential. The induced potential is

$U = Blv$

where $B$ is the magnetic field at that position, and $l$ is the length of wire and $v$ is the velocity the edge cutting magnetic field. But these two induced potential are having opposite directions, and the two edges are moving at the same velocity ($10m/s$), so the total $EMF$ induced should be

$U = B_{r=2cm}\cdot a v - B_{r=3cm}\cdot a v$

-
Thank you so much. The explanation was a little confusing, but I managed to figure it out. I believe EMP should be EMF in the last sentence there. This will get the correct answer :D –  Charles Nov 20 '11 at 17:19