ElectroMagnetic Induction - HELP

[krugertown]

A 20cm x 20cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis. The resistance of the loop is 0.50$$\Omega$$. A uniform magnetic field parallell to the z-axis is given by B=0.80y$$^{2}$$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 = 0.50s?
Induciton of Solonoid = ((magnetic constant)N^2A)/I
$$\epsilon$$ = d$$\phi$$/dt
$$\phi$$ = magnetic fluxi thought first i would need to differentiate the original equation to find the emf of the loop and hence the current.
but that's where i got stuck? where does the y value come in?

once I've got the current i can find the induced current but am struggling with how to get the emf of the loop!

help would be appreciated!

 

[marcusl]

The B field is non-uniform and it's value depends on y. You will have to integrate it to find the enclosed flux. Your second equation gives the emf (although you left off the negative sign). That and the resistance give the current.

 

[thomate1]

Hello,

Thank you for reaching out for assistance with this problem. I can provide some guidance and clarification on how to approach this question.

Firstly, it is important to understand the concept of electromagnetic induction. This is the production of an electromotive force (emf) in a conductor when it is exposed to a changing magnetic field. In this case, the square loop of wire is the conductor and the changing magnetic field is given by the equation B=0.80y^{2}t.

To find the induced current in the loop at t=0.50s, we need to use the equation for induced emf, which is given by \epsilon = -N\frac{d\phi}{dt}, where N is the number of turns in the loop and \phi is the magnetic flux. In this case, we have a single turn in the loop, so N = 1.

The magnetic flux, \phi, is the product of the magnetic field strength and the area of the loop. Since the loop is in the xy-plane and the magnetic field is parallel to the z-axis, we can find the area of the loop by taking the product of its length and width, which is 20cm x 20cm = 0.04m^2.

Now, we need to find the magnetic field strength at t=0.50s. To do this, we plug in t=0.50s into the given equation for B. This gives us B=0.80(0.50)^{2}(0.50) = 0.10T.

Substituting the values we have found into the equation for induced emf, we get \epsilon = -(1)(0.10)(0.04) = -0.004V.

Since the resistance of the loop is given as 0.50\Omega, we can use Ohm's law (V=IR) to find the induced current, which is I=\frac{\epsilon}{R} = \frac{-0.004}{0.50} = -0.008A.

Therefore, the size of the induced current in the loop at t=0.50s is 0.008A. It is important to note that the negative sign indicates that the current is flowing in the opposite direction to the magnetic field.

I hope this explanation helps you to understand the concept of electromagnetic induction and how to apply it to this