Question

Two current carrying wires carry currents of 4A and 3A. Calculate the magnetic field at point P

Answer 1

currents in the wires , 1 and 2 is

I₁ = 4 A

  I₂ = 3A

distance between the wires , d = 2 cm

                                                   = 2 cm (10^-2 m /1 cm )

                                                    = 0.02 m

distance b/w current I₁ to point P₁ is ,

                                               r ₁ = 1 cm

                                                     = 1 cm (10^-2 m /1 cm )

                                                    = 0.01 m

distance b/w current I₂ to point P₃ is ,

                                               r ₃ = 1 cm

                                                    = 1 cm (10^-2 m /1 cm )

                                                    = 0.01 m

distance b/w I₁ or I₂ to piont P₂ is ,

                                              r ₂ = 2/2 cm

                                                    = 1 cm (10^-2 m /1 cm )

                                                    = 0.01 m

         velocity of the electron . v =  4 x 10⁵ m/s

magnetic field (magnitude and direction) at points, P1, P2 and P3 is ,

   at : P₁ is , B₁ = ?o I₁ / 2?r₁

                         = (4?x10^-7 A/m )(4 A) / 2?(0.01 m)

                         = 800 x10^-7 T (in to page )    at : P₂ is ,

B₂ = ?o I₁ / 2?r₁ -  ?o I₂ / 2?r₁     =  800 x10^-7 T - 600 x10^-7 T      = 200 x10^-7 T (out of page)at

: P₃ is ,B₃    =?o I3/2pi = 800 x10^-7 T (in to page)