Saved Bookmarks
| 1. |
There is one solenoid of length I and self-inductance L. What is the length of wire used to make this solenoid? |
|
Answer» `sqrt((pilL)/(mu_(0)))` `phi = (mu_(0)NI)(pir^(2))(nl)` Comparing it with `phi = LI`, we get self-inductance as follows: `L= mu_(0)pin^(2)r^(2)l` Total length of wire USED to make solenoid can be written as follows: `L.= (2pir)(nl)= 2pirnl` Value of nr to be substituted in equation (i) can be TAKEN from equation (i) as follows: `n^(2)r^(2)= (L)/(mu_(0)pil)` `nr= sqrt((L)/(mu_(0)pil))` Substituting in equation (ii) we get length of wire as follows: `L.= 2pi sqrt((L)/(mu_(0)pil))` `L.= sqrt((4pilL)/(mu_(0)))` |
|