Respuesta :
The top page will receive the current in the loop, and the bottom page will receive it out of the loop.
(a) Because B is constant and there is a constant magnetic Feld, we may recast the change in flux as shown above. Since A is positive, the loop will have a negative emf, which corresponds to an induced magnetic moment on the page that points to the left. The loop's current will flow into and out of the page in opposite directions. V=-Δφ/ Δt = -B·ΔA/ Δt
(b) Individual electrons in the wire loop can be imagined travelling radially outward as the loop's radius grows. We'll take into account one at the top of the loop (which is moving up the page). The Lorentz force Law is F=q(v)=qvB(yx)=-qvB.
(c) Circular orbits will emerge from constant forces pointing into the center of the loop (around the wire). We can tell we have positive current there since the force is pointing into the middle of the loop (into the page).
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