Lost with the whole question! lots of help would be appreciated An electromagnet
ID: 1930246 • Letter: L
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Lost with the whole question! lots of help would be appreciated
An electromagnet generates a B field with x,y, and z components (Asin(omega t), 0, Asin(omega t), where A is constant. What is the normal vector of the resulting E fields? Prove this using Faraday s Law in differential form. |The field penetrates a disk of diameter D that is centered on position (0,0,0). In what direction should the surface normal of the disk be in order that the EMF around its perimeter is a maximum? Assuming that all E field components are zero at x=y=z=0, sketch the electric field distribution around the origin at time t= 0 in the plane of the disk. You may get an idea for a trial form of solution from question 3 (shifted to the appropriate plane and change sin to cos). Show that the resulting field strength is proportional to r, the distance from the origin. Integrate the field around a radius, r, to find the EMF that could be generated if there were a loop at that radius. Your result should be consistent with Faraday's Law in integral form. Is it best to make the loop small or large in order to maximize voltage? If A=B0 is 1 Tesla, and a loop is placed at a radius of R=1 m, what is the voltage amplitude generated in a single loop at 60 Hz? If there are ten loops at that radius what voltage can be achieved?Explanation / Answer
curl(E)=-dB/dt = -Awcos(wt)i-Awcos(wt)k where i and k are unit vectors along the x and z directions
By defn of curl, dEz/dy-dEy/dz=-Awcos(wt), dEx/dz=dEz/dx, and dEy/dx-dEx/dy=-Awcos(wt).
Now for some visualization. The curl of E, which is parallel to B as seen above, lies in the x-z plane, and is at 45 degrees to x and z axes. The lines of E will then be in a plane perpendicular to B. A guess solution is that the E field have a spatially constant component (with a time-dependence, perhaps) along the x and z axes. Thus all partical derivatives wrt x and z are 0, and you get two simple equations for Ey: -dEy/dz=-wcos(wt), and dEy/dx=-wcos(wt). These are the simplest possible differential equations. Thus apart from a constant that might depend on time and y, Ey=A(-x+z)wcos(wt)
The solution is E = f(t)i+[A(-x+z)wcos(wt)+g(t,y)]j+h(t)k. This was our first guess. The most general solution consistent with Faraday's law can now be written
E = f(t,x)i+[A(-x+z)wcos(wt)+g(t,y)]j+h(t,z)k. I cannot think of any legal simplification, although I'd recommend that you state the answer simply as E = [A(-x+z)wcos(wt)]j, since this is the simplest answer consistent with Faraday'slaw.
(b) The disc must be perpendicular to the magnetic field for the flux through it to be a max, since the emf along the perimeter is proportional to the (time-derivative of) the flux through the disk. Disc is perp to the B field, meaning, its normal is along the B field, B=Asin(wt)(i+k). So the answer is that the nornal to the disc must be along the unit vector (i+k)/2.
(c) ----> ---> --> . <-- <--- <----
Flip each arrow vertically, and repeat the pattern above and below.
The dot represents the center (0,0,0), and indicate that the unit normal to the plane is (i+k)/2.
No idea on (d) and (e). Sorry.
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