Saved Bookmarks
| 1. |
Obtain an expression for the magnetic force on a current carrying conductor. |
|
Answer» Solution :Let l be the length of a conductor. Let A be its area of cross section. Let n be the number density of mobile charge carriers in the conductor. The total number of mobile charge carries is NAL. Let the averrage drift velocity be `'v_(d)'` corresponding to the current I. Let B be an external magnetic field surrounding the conductor. Magnetic force on these carriers is, `vec(F)=(nAlq)(vec(v_(d)) times vec(B))` where `q^(')=nalq` By DEFINITION current density `j=I/A` and q = e (charge on the ELECTRON) `""vec(F)="nAle "vec(v)_(d) timesvec(B)` `""=("NE"vec(v)_(d))Al times vec(B)` `""=vec(j)Al times vec(B)` `""=abs(vec(j))(HAT(j)l)A times vec( B)` `therefore""vec(F)=Ivec(l) times vec(B)` The direction of current is in the direction of `vec(l)` vector, `vec(l)=hat(j)l " and " jA=I=` current. 
|
|