EMF in a square loop due to magnetic field from an infinite wire

[Loopas]

Homework Statement

Determine the emf induced in the square loop in the figure if the loop stays at rest and the current in the straight wire is given by I(t)=(15.0A)sin(2500t) where t is in seconds. The distance a is 12.0 cm, and b is 15.0 cm.

Homework Equations

emf = Δmagneticflux/Δt
magnetic flux = BA
magnetic field from infinite wire = (μI)/(2*pi*R)

The Attempt at a Solution

I tried using the using the magnetic field from the wire and the area of the square loop to calculate magnetic flux. My final answer was: (2.9*10^-7*sin(2500t))/t

But this isn't the right answer, so I'm guessing that I have to integrate with Faraday's Law but I am clueless about integrals

 

[TSny]

Hello, Loopas.

so I'm guessing that I have to integrate with Faraday's Law

That's right.

but I am clueless about integrals

Does that mean you don't know how to evaluate elementary integrals? Is this problem from a course that requires calculus as a prerequisite?

 

[jaytech]

Also, the EMF = - $\frac{dø}{dt}$, meaning you need to take a derivative with respect to t, not divide by t.

 

[Loopas]

Well let's put it this way... I'm supposed to have a clue about integrals, but it's something that still eludes my full understanding, especially when integrating things like Faraday's Law, Ampere's Law, all of that good stuff.

So I'm guessing that the first step in this problem should be to integrate the magnetic field of b > r > b+a using Ampere's Law? And then Faraday's Law is somehow used find the emf?

 

[TSny]

Yes. You'll first need to find the total flux through the loop. Since the field varies with distance from the wire, you cannot use $\small \Phi = BA$. Instead, you will need $\small \int{BdA}$

The picture has already helped you by showing a thin strip of thickness dr located at a distance r from the wire. What is an expression for B at the location of this strip in terms of $\small r$? What is an expression for the area $\small dA$ of the strip in terms of $\small a$ and $\small dr$?

 

[Loopas]

Ok, so $\int\frac{μ*I(t)}{2*pi*r}*adr$

And this will be integrated from b to b+a?

 

[TSny]

Yes, that looks good. Once you have the total flux $\small \Phi$ you can use Faraday's law to calculate the emf.

 

[Loopas]

I just want to make sure I'm doing this right --

I found the integral to be $\frac{μI(t)a}{2pi}$(ln(2pir))

Integrating from a to a+b it would be --

$\frac{μI(t)a}{2pi}$(ln(2pia+2pib)-ln(2pia))

 

[Loopas]

Actually I messed up before, this should be correct I think

$\frac{μ*I(t)*a}{2*pi}$(ln(a+b)-ln(a))

 

[Loopas]

Finally found it!

emf = -(5.3*10^-4)(cos(2500t))

Thanks for the help!

 

[TSny]

Great! Good work.