Saved Bookmarks
| 1. |
An iron rod, 2 cm in diameter and 20 cm long is bent into a closed ring and is wound with 3000 turns of wire. It is found that when a current of 0.5 A is passed through this coil, the flux density in the coil is 0.5 Wb/m2 . Assuming that all the flux is linked with every turn of the coil, what is (a) the B/H ratio for the iron (b) the inductance of the coil ? What voltage would be developed across the coil if the current through the coil is interrupted and the flux in the iron falls to 10 % of its former value in 0.001 second ? |
|
Answer» H = NI/l = 3000 × 0.2 = 7500 AT/m B = 0.5 Wb/m2 (a) Now, B/H = 0.5/7500 = 6.67 x 10-5 H/m. Also μr = B/μa H = 6.67 × 10−5 /4π × 10−7 = 53 (b) L = NΦ/I = (3000 x π (0.02)2 x 0.5)/(4 x 0.5) = 0.94H eL = N(NΦ/dt) volt; dΦ = 90 % of original flux = (0.9 x π (0.02)2 x 0.5)/4 = 0.45π x 10-4Wb dt = 0.001 second ∴ eL = 3000 × 0.45π × 10−4 /0.001 = 424 V |
|