ElectroMagnetic Induction - HELP

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To determine the induced current in a 20cm x 20cm square loop of wire in a non-uniform magnetic field, the magnetic flux must be integrated due to the field's dependence on the variable y. The induced electromotive force (emf) can be calculated using the formula ε = dφ/dt, where φ represents the magnetic flux. The resistance of the loop, given as 0.50Ω, is essential for calculating the current once the emf is determined. The negative sign in the emf equation indicates the direction of the induced current according to Lenz's Law. Overall, the integration of the magnetic field and the application of the resistance will yield the size of the induced current at t = 0.50s.
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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!
 
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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.
 
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