Pratham Ashish
Last Activity: 15 Years ago
hi ,
first the magnetic field at any point by i1 willbe = µ/2pi *i1 / r
force on any current element = i2 * ( dl X B)
let i2 is placed at a distance apart from origin
by vector anaysis force on upper part of I2 will be in +z direction whereas on lower part it willbe in -z direction
let any current element dl substed Ø at the origin with positive x direction then
l/a = tan Ø
emplies dl = a sec^2 Ø dØ
& r = a secc
putting in (1)..
dF = i2 * ( a sec^2 Ø dØ * µ/2pi *i1 / a sec Ø ) sin Ø
= µ/2pi * I1*I2 * tan Ø dØ
total force on upper part ,
F1 = §[ µ/2pi * I1*I2 * tan Ø dØ ]030
on lower part F2 = §[ µ/2pi * I1*I2 * tan Ø dØ ]060
total force on wire = F1 -F2
= - §[ µ/2pi * I1*I2 * tan Ø dØ ] 030
= µ/2pi * I1*I2 *ln sec Ø ] 030
=µ/2pi * I1*I2 *ln 2/3 1/2 in the -z direction