Question

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)?

Answer 1

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=$\sqrt{b1^2+b2^2}$

the magnetic field due this long wire is given by

B1=∨I1/$(2\pi *R1)$..............................1

B2=∨I2/$(2\pi *R2)$............................2

Bnet=$\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}$.......................3

Bnet=v/2*pi$\sqrt{(I1/R1)^2+(i2/R2)^2}$

Bnet=4*pi*10^-7/(2$\pi$)$\sqrt{(43/1.7)^2+(41/29.5)^2}$

Bnet=0.0000002*(641.72)^.5

Bnet=1.006*10^-6T