Answer:
Bnet=1.006*10^-6T
Explanation:
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?
the magnetic field Bnet=
the magnetic field due this long wire is given by
B1=∨I1/..............................1
B2=∨I2/............................2
Bnet=.......................3
Bnet=v/2*pi
Bnet=4*pi*10^-7/(2)
Bnet=0.0000002*(641.72)^.5
Bnet=1.006*10^-6T