u2 YgPg

QrHH$

B-1 BY (u + Va )Pc -KPC)) - (Yc - 1)(u + Va )pC = 0. (3.14.3) Here the gravitational potential f is taken as f = fo (1 + s/s1 P1, (3.14.4)

where fo = 1.9x 10 cm /sec , S1 = 45 kpc . The shape of flux tube is taken in the form

B(s) = Bo(1 + (s/s2P) r(s)= ro(1 + (s/s2P)/2, (3.14.5)

where s2= 15 kpc. The calculations are made at yg = 1.6 for the boundary condition at zo = 3 kpc as follows:

Was this article helpful?

## Post a comment