Saved Bookmarks
| 1. |
Obtain an expression for magnetic field inside the toroid. |
|
Answer» <P> Solution :CONSIDER three circular Amperian loops 1, 2 & 3 as shown in the figure by dotted line.Let the magnetic field along loop 1 be `B_(1)` in magnitude. The length of the length `intdl=2pir_(1)`. This current loop does not enclose current. Therefore `I_(e)=0 " and " B_(1)=0`. Therefore, the magnetic field at any POINT P in the open space inside the toroid is zero. Consider the Amperian loop 3. Let `B_(3)` be the magnetic field along the loop. Here `intdl=2pir_(3)`. The current coming out of the planes is cancelled exactly by the current going into it. Therefore `I_(e)=0 " and " B_(3)` at any point Q outside the toroid is zero. Let the magnetic field inside the toroid be B. The length of Amperian in loop `2=intdl=2pir_(2)`. The current enclosed for N turns of toroidal COIL is NI. From Amperian Circuital law, `""Bintdl=mu_(0)I_("enclosed")` i.e., `""B(2pir_(2))=mu_(0)NI` i.e., `""B=(mu_(0) NI)/(2pir_(2))` Let the number of turns per unit length be n. Then `N/(2pir_(2))=n`. Thus, `""B=mu_(0)nI`. 
|
|