InterviewSolution
Saved Bookmarks
| 1. |
A metal rod of resistance of 15 Ω is moved to the right at a constant 60 cm/s along two parallel conducting rails-25 cm apart and shorted at one end. A magnetic field of magnitude 0.35 T points into the page, (a) What are the induced emf and current in the rod? (b) At what rate is thermal energy generated? |
|
Answer» Data: R = 15Ω, v = 0.6 m/s, l = 0.25m, B = 0.35T (a) Induced emf, e = Blv = (0.35)(0.25)(0.6) = 0.0525 V = 52.5 mV The current in the rod, I = \(\cfrac{e}R\) = \(\cfrac{52.5}{15}\) = 3 5 mA (b) Power dissipated, P = eI = 0.0525 × 3.5 × 10-4 = 0.184 mW |
|