Saved Bookmarks
| 1. |
Two horizontal wires PQ and RS of resistance 10 Omega and 20 Omega are separated by a distance or 10 cm and connected in parallel in vertical plane across a cell of emf 200 V and negligible internal resistance. A wire AB of mass 1 g and length 1 cm in balanced exactly,midway between them. What must be the current in it ? |
|
Answer» Solution :AB experience a force of attraction force PQ and RS. Apply V = iR ` 200 i_1 xx 10 RARR i_1 = 20 A` `200 =I_2 xx 20 rArr i_1 = 10 A` AB will be in equilibrium if `(mu_0)/( 2pia) i_2 i_3 l + MG = (mu_0 i_1 i_3)/(2 pi a)` `(mu_0)/(2pi a) xx (i_1 i_3 - i_2i_3) l = mg `  `(4 pi xx 10^(-7))/(2 pi xx 5 xx 10^(-2))(20 i_3 -10 i_3) xx (1)/(100) =10^(-3) xx 9.8` ` (2 xx 10^(-5) xx 10 i_3)/(5) = 9.8 xx 10^(-3)` `i_3 = 245A.` |
|