Question

4. Two long, parallel wires carry currents of 11 = 3.18 A and I2 = 4.50 A in the direction indicated in the figure below. (Choose the line running from wire 1 to wire 2 as the positive x-direction and the line running upward as the positive y-direction.) Find the magnitude and direction of the magnetic field at point P. located d = 20.0 cm above the wire carrying the 4.50-A current. 1 12 -d

Answer 1

I1 = 3.18A  

I2 = 4.50 A

d = distance along x-axis between wires = 0.20 m

P ( 0.20m , 0.20m )

r1 = sqrt(2) d ,

B1 = 4pi*10^(-7) H/m*3.18 A/ (2*pi*sqrt(2)*0.20m) = 2.25*10^(-6) T ( direction 135 deg with ref to +x-axis )

r2 = 0.20 m

B2 = 4pi*10^(-7) H/m*4.50 A/ (2*pi*0.20m) = 4.5*10^(-6) T ( direction 180 deg with ref to +x-axis )

Resultant = 6.30*10^(-6) T , 165 deg with respect to +x-axis

[vector additon : x-axis = (-4.5 - 2..25 cos45 ) *10^(-6) = - 6.09 *10^(-6) T

y-axis = 2.25*sin(45) = 1.59 *10^(-6) T

resultant = sqrt(6.09^2+1.59^2)*10^(-6) = 6.30*10^(-6)

angle = arc tan ( 1.59/-6.09) = -15 deg with ref to -x-axis or 165 deg with respect to x-axis]