Induction on a Loop Homework: 0.520Ω Square Wire

[theowne]

Homework Statement

A 20.0 cm by 20.0 cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis. The resistance of the loop is 0.520 {\Omega}. A magnetic field parallel to the z-axis is given by B = 0.790 * y^2 * t, where B is in tesla, y in meters, and t in seconds. What is the size of the induced current in the loop at t = 0.550 s?

Homework Equations

flux = A.B.cos theta

The Attempt at a Solution

I'm having a hard time understanding this question. I see that cos theta would be one so I can use the formula A B for the flux. But what is the y^2? Is it a variable? How can I construct the d(flux) / dt? I know that the A can be pulled out, and the 0.790 as well, but I'm not sure how to treat the other two.

 

[Redbelly98]

Yes, y is a variable. Calculating the flux involves doing an integral.

 

[theowne]

Okay but what variable is it? And why does calculating the flux involve doing an integral?

 

[Redbelly98]

A 20.0 cm by 20.0 cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis.

Do you understand what the xy plane is? x and y are just what they usually are. They are the x and y coordinates in the xy plane.

... why does calculating the flux involve doing an integral?

The definition of flux is an integral:
http://en.wikipedia.org/wiki/Magnetic_flux#Description

 

[theowne]

Yes, somehow I can grasp what an x y plane is, but what I meant was for clarification was what it was the y coordinate of. Is it saying the field is different at different y coordinates. And the only definition of flux we have been given to use is area cross magnetic field.

 

[tjkubo]

Is it saying the field is different at different y coordinates. And the only definition of flux we have been given to use is area cross magnetic field.

Yes, you are exactly right. However, since B is a function of y, you cannot just take out the area from the integral (the flux is not uniform over the entire square loop). You must divide the square into small area differentials dA.

 

[bowma166]

Yes, as tjkubo said, for this problem you can't simply say

$$\Phi_B=BA\cos\theta$$

but instead must use

$$\Phi_B=\int\vec{B}\cdot\textrm{d}\vec{A}$$

How would you write $\textrm{d}\vec{A}$ in terms of dy in this problem?