Saved Bookmarks
| 1. |
An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R ) respectively, in a uniform magnetic field vecB=B_(0)hati, each with an equal momentum of magnitude p = eBR. Under what conditions on the direction of momentum will the orbits be non-intersecting circles ? |
Answer» Solution :1. If RADIUS of the circle made from combination of electron and positron pair then if this is more than 2R, than these two circle does not intersect with each other.  2. Magnetic field `vecB` is in the direction of x, so momentum of circular system is in plane yz. 3. Both particles are opposite to each other from orbital radius R. 4. Let `P_(1)andP_(2)`, are momentum of electron and positron. 5. Suppose `P_(1)` form `theta` angle with y-axis and `P_(2)` also form same angle. 6. Centre of RESPECTIVE circle are at origin position and perpendicular to R distance. 7. Centre point of path of electron-positron circle as shown by `C_(E)andC_(p)`. 8. Coordinate of `C_(e)(0,-Rsintheta,Rcostheta)` co-ordinate of `C_(p)(0,-Rsintheta,3/2R-Rcostheta)` 9. If distance between centre point to radius for circle is GREATER than 2R than circle doesn.t inter sect with each other. 10. Suppose, distance between `C_(p)andC_(e)` be d. `therefored^(2)=(2Rsintheta)^(2)+(3/2R-2Rcos0^(@))^(2)` `d^(2)=4R^(2)sin^(2)theta+9/4R^(2)-6R^(2)costheta+4R^(2)cos^(2)theta` = `4R^(2)+9/4R^(2)-6R^(2)costheta` 11. If distance of d be more that 2R, `dgt2R` `d^(2)gt4R^(2)` `therefore4R^(2)+9/4R^(2)-6R^(2)costhetagt4R^(2)` `therefore9/4R^(2)gt6R^(2)costheta` `therefore9/4gt6costheta` `thereforecosthetalt3/8` So, this is necessary condition. |
|