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An infinite wire , placed along z axis has current I1 in positive z direction . A conducting rod placed parallel to y axis has current I2 in positive y direction. The ends of rod subtens +30 and -60(degrees) at the origin with positive x direction.The rod is distance a from the origin . Find net force on the rod. An infinite wire , placed along z axis has current I1 in positive z direction . A conducting rod placed parallel to y axis has current I2 in positive y direction. The ends of rod subtens +30 and -60(degrees) at the origin with positive x direction.The rod is distance a from the origin . Find net force on the rod.
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
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
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