Question

A square filamentary loop of wire is 25 cm on a side and has a resistance of 125 Ohms per meter length. The loop lies in the z = 0 plane with its corners at (0,0,0), (0.25,0,0), (0.25,0.25,0), and (0,0.25,0) at t = 0. The loop is moving with velocity vy = 50 m/s in the field Bz = 8 cos (1.5×108t −0.5x) µT. Derive an expression for the power as a function of time being delivered to the loop.

Answer 1

The flux at the original loop maintains: 0.25 0.25 $(t)= 5 | Bydxdy 0 0 0.25 Bed The plane is not moving y direction 0 0.25 0.25 = = [ (8x10^)cos(1.5x10°7–0.5x)di (8x10^^)[-sin (1.5x10°t – 0.5x)]”* (0.5) =(-4x10")[sin (1.5x10°t – 0.5(0.25)) – sin (1.5x10ʻt – 0.5(0.5))] =(-4x10~)[sin (1.5x10°+-0.125) – sin (1.5x10°r)] Using Faraday's law: = 4x10-6 do v(t)=-4 -- [ -4*10* sin(1.5x10°t -0.13)–sin (1.5x10° +1)] sin (1.5x10°1–0.13) - sin(1.5x108+1) = 4x 10[1.5*10* cos(1.5x10ʻt-0.13)-1.5*10* cos(1.5x10$+1)] = (4x10-)(1.5*10*)( cos(1.5x10ʻt-0.13) - cos(1.5x10°t)] = (6x102)[cos (1.5*10*t0.13) - cos (1.5x10ºr)] d dt Time varying power: v2(t) P(t)= R (6x102)* [cos(1.5x10°t - 0.13) – cos(1.5x10't)] 125 = (2.9x10°)[cos(1.5x10°t -0.13) – cos(1.5*10°r)]