Two long, parallel conductors, separated by 14.0 cm, carry currents in the same

Question

Two long, parallel conductors, separated by 14.0 cm, carry currents in the same direction. The first wire carries a current I_1 = 2.00 A, and the second carries I_2 = 8.00 A. (See figure below. Assume the conductors lie in the plane of the page.) What is the magnetic field created by 2_1 at the location of I_2? magnitude T Direction What is the force per unit length exerted by I_1 on I_2? magnitude N/m direction What is the magnetic field created by I_2 at the location of I_1? magnitude T Direction What is the force per length exerted by I_2 on I_1? magnitude N/m direction

Explanation / Answer

Given

   two wires are of carrying current i1 = 2 A, i2 = 8 A are

   separated by r = 0.14 m

we know that the magnetic field at a distance r from a current carrying wire is

      

   B = (mue0*i/2pi*r)

a) magnetic field at i2 due to i1 is

   B1 = (mue0*i1/2pi*r) = (4pi*10^-7*2/(2pi*0.14)) T

   = 2.857143*10^-6 T

direction is into the plane

b) magnetic force per unit length exerted by i1 on i2 is
F = BiL sin theta here theta is 90

   F12/l = (mue0*i1*i2/2pi*r)

       = (4pi*10^-7*2*8/(2pi*0.14)) N/m

       = 2.2857142857143*10^-5 N/m

       = 22.857143*10^-6 N/m
the force is downward (by right hand rule)

c) magnetic field at i1 due to i2 is

   B1 = (mue0*i2/2pi*r) = (4pi*10^-7*8/(2pi*0.14)) T

   = 11.42857143*10^-6 T

direction is into the plane

d) magnetic force per unit length exerted by i1 on i2 is
F = BiL sin theta here theta is 90

   F12/l = (mue0*i1*i2/2pi*r)

      
       = (4pi*10^-7*2*8/(2pi*0.14)) N/m

       = 2.2857142857143*10^-5 N/m

       = 22.857143*10^-6 N/m

the force is downward (by right hand rule)

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