A rod of length L moves at a constant velocity V, along conducting horizontal ra

Question

A rod of length L moves at a constant velocity V, along conducting horizontal rails as shown in Figure below. The magnetic field thought the conducting loop is provided by a very long parallel wire carrying a constant current I. Assume that V, a, L, and I are all given. (a)Calculate the resistant, magnitude and the direction of the induced current in the conducting loop. How much work is done in moving the rod L0 meter to the left? Calculate the induced voltage epsilon using the induced electric field E_ind.

Explanation / Answer

The magnetic field at a distance x from an long wire is given as: B = oI / 2x

Now, let us consider a thin strip of the area inside the loop of thick dx and distance x from the wire. We will also assume that the moving rod is at distance y from the right side end of the loop.

So, the magnetic flux through the small width dx would be: d = Bydx where B = oI / 2x

That is, the net magnetic flux would be: = d = oIydx / 2x = [oIy/2] ln[(L+a)/a)

Now, the rate of change of the magnetic flux would be equal to the electric potential induced in the loop.

That is d / dt = [oI/2] ln[(L+a)/a) dy/dt = [oIV/2] ln[(L+a)/a)

Part A.) The current induced in the loop would be EMF / Resistance = [oIV/2R] ln[(L+a)/a)

Part B.) Now the force acting on the rod of length Lo with current [oIV/2R] ln[(L+a)/a) and magnetic field B, would be: ILB = [V/R][oI/2]2 [ln[(L+a)/a)]2 [The magnetic field would vary from the one end to the other and hence needs to integrated the way magnetic flux was done.]

Part C.) The voltage induced as calculated above would be: [oIV/2] ln[(L+a)/a)

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