Induced Emf in Rectangular Loop Problem

[ethanabaker1]

Homework Statement

"You push a rectangular loop of width .2m, length .8 m, and resistance R=200$\Omega$ into a magnetic field (out of page) B_out=.4 T and a speed v=.2 m/s. The long side of the rectangle is parallel to the x axis. What is the current flowing in the loop?"

Homework Equations

I think I should use:
Emf=vBl
Ohm's Law - $\Delta$V=IR

The Attempt at a Solution

Using the equation emf=vBl, I did emf=(.02)(.8)(.4)=.0064.
Then I said that emf=$\Delta$V, so using Ohm's Law I did $\frac{.0064}{200\Omega}$ and got 3.2 x 10^-5 A as my answer.

My problem is that the only problems we've worked are for circular loops and I'm not sure what length to use for l.

 

[Andrew Mason]

Homework Statement

"You push a rectangular loop of width .2m, length .8 m, and resistance R=200$\Omega$ into a magnetic field (out of page) B_out=.4 T and a speed v=.2 m/s. The long side of the rectangle is parallel to the x axis. What is the current flowing in the loop?"

Homework Equations

I think I should use:
Emf=vBl
Ohm's Law - $\Delta$V=IR

The Attempt at a Solution

Using the equation emf=vBl, I did emf=(.02)(.8)(.4)=.0064.
Then I said that emf=$\Delta$V, so using Ohm's Law I did $\frac{.0064}{200\Omega}$ and got 3.2 x 10^-5 A as my answer.

My problem is that the only problems we've worked are for circular loops and I'm not sure what length to use for l.

Your equation is derived from Faraday's law. So apply Faraday's law explicitly:

$$emf = \oint \vec{E}\cdot d\vec{s} = \frac{d\phi}{dt} = B\frac{dA}{dt}$$

All you have to do is work out dA/dt, the rate of change of the area enclosed by the loop. If l is the distance along the x-axis through which you have pushed the loop and w is the width of the loop what is dA/dt?

AM

 

[ethanabaker1]

How would I do this algebraically?

 

[Andrew Mason]

How would I do this algebraically?

Since dA = wdx and w is constant, then dA/dt = w(dx/dt) = wv

So emf = Bwv

AM

 

[rwisz]

Your attempt at a solution is correct. The length l in the equation emf=vBl refers to the length of the conductor in the magnetic field, which in this case is the long side of the rectangular loop. So your calculation of emf is correct, and then using Ohm's Law to find the current flowing in the loop is also correct. Good job!