Inductance of a loop in a nonuniform magnetic field

[sfdevil]

Homework Statement

A square loop made of wire with negligible resistance is placed on a horizontal frictionless table. The mass of the loop is m and the length of each side is b. a nonuniform vertical magnetic field B=B0(1+kx) exists in the region, where B0 and k are constants. The loo is given a quick push with initial velocity v along x-axis. The loop stops after a time interval T. Find the inductance of the loop.

Homework Equations

emf(ind) = -L*dI/dT
U=1/2*L*I^2
emf= -delta flux/delta t

The Attempt at a Solution

well, I am sort of in a loss for this one.
I tried to get the induced emf by finding dflux/dt:

flux
=integral ( B0*(1+kx) * b dX )
= b*B0*(b*2*k*x+b^2*k+2*b)/2

change of flux in regards with time = dphi/dx * dx/dt
= b^2*k*B0*v (because velocity= dx/dt)

so emf is b^2*k*B0

now I am not sure what to do, since the resistance is negligble and i can't find the current from it...

also the whole time interval thing, where does it come into play (kinematics ?)
and should I use conservation of energy here ?
1/2 * m * v^2 = 1/2 * L * I^2 ? for some reason I don't think its the right way

just thoroughly confused with this one. please help !

 

[Jhero]

Thank you for your post. I understand your confusion with this problem. Let me try to guide you through the solution.

First, let's start with the basics. The inductance of a loop is given by the formula L = N*ϕ/I, where N is the number of turns in the loop, ϕ is the magnetic flux through the loop, and I is the current in the loop. In this case, we have a single-turn loop, so N = 1.

Next, let's consider the given information. The loop is given a quick push with initial velocity v along the x-axis. This means that the loop will experience a force in the y-direction due to the nonuniform magnetic field. This force will cause the loop to rotate, and eventually come to a stop after a time interval T. This information is important because it tells us that there is a change in the magnetic flux through the loop over time.

Now, let's look at the equation for induced emf that you have correctly identified: emf(ind) = -L*dI/dT. This equation tells us that the induced emf in the loop is equal to the negative of the inductance multiplied by the rate of change of current in the loop. In this case, we are interested in finding the inductance of the loop, so we need to find a way to calculate the current in the loop.

To find the current in the loop, we can use Kirchhoff's voltage law. This law states that the sum of the emfs in a closed loop must be equal to the sum of the potential differences in that loop. In this case, the only emf in the loop is the induced emf, and the only potential difference is the potential difference across the loop due to the nonuniform magnetic field. This potential difference can be calculated using the formula V = B*dL, where B is the magnetic field and dL is the length of the loop. Since the magnetic field is changing with position, we need to integrate this formula over the length of the loop. This gives us:

V = integral (B0*(1+kx)*b dx) = B0*b*(b*k*x + b^2*k + 2*b)/2

Now, using Kirchhoff's voltage law, we can set this potential difference equal to the induced emf and solve for the current:

emf(ind) = V =