Question

Recall that it is conservative, then the line intera/ F.dr is path-independent meaning that the the integral depends only on the initial and terminal bolets of the sath, and not on the path Similar ideas are true for surfaces, although we must now discuss the curl instead of the gradient. Note that there is some vector field A such that (V x A) = F. then Suo ' Theorem tells us that JP as - x A). S = 6 A-de This vector field A is called a vector potential for F. Note that the value on the furthest right integral does not depend on S - only on its boundary. Hence, we might say that the surface integral of any vector field with a vector potential is surface-independent, and its value depends only on the line integral of the potential around the boundary For the next example, consider a uniform magnetic field B that has a constant strength in the direction meaning that B = (0,0,b) for some parameter b. () (1 Credit) If r is the vector field ( 7.0), verify directly that A B X is a vector potential for B by computing its curl. (b) (1 Credit) Now, suppose we want to compute the flux of Bupward through S, the unit hemisphere VI- y. Use Stokes Theorem to compute this flux by computing A dr. This shows us that computing a vector potential is helpful for evaluating surface integrals, just like a scalar potential before it. However, algebraically finding a vector potential is tough-physicists will often rely on

Answer 1

IXA= N (1) A = L Bxr RXA B= (0,0, b) r= (x, y, o). Bxr = 0 o 6 ay byt xb} = (-yb, xb, o). PXA2 % toy yoz = Col +of(6+b))= 60,00 Tyb ab ol (6) flus - Andr - 1 (yb, xb,o) .dr. as, as. dif - dr r + rhod + rsino do §. =/doệt sinod & & Xr=/ B. dr = dxê + dy i+dzi so flux = 2 gay bdx + 1 fabdy x² + y² + z ² =4. = I bf - V1 = x² dx + 1 of Ji-yody = 0 asi