Question

One long wire lies along an x axis and carries a current of 39 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 4.4 m, 0), and carries a current of 47 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 0.80 m, 0)

Answer 1

Answer:

Explanation:

Magnetic field due to a long current carrying wire can be calculated as follows .

B = 10⁻⁷ x 2I / d where B is magnetic field , I is current .

The wire is along x -axis and the point is on y-axis at a distance of 0.8 m

Magnetic field at point of .8 m on y -axis

B₁ = 10⁻⁷ x 2 x 39 / 0.8  

= 97.5 x 10⁻⁷ T .

Second wire is parallel to  z-axis and passes through point on y-axis at a distance of 4.4 m . So the given point is at a distance of 4.4 - .8 = 3.6 m

Magnetic field

B₂ = 10⁻⁷ x 2 x 47 / 3.6  

= 26.11  x 10⁻⁷ T .

Both these magnetic fields are perpendicular to each other so

Resultant magnetic field

B = √ ( 26.11² + 97.5² ) x 10⁻⁷ T

= √( 681.73 + 9506.25 ) x 10⁻⁷ T

= √( 10187.98) x 10⁻⁷ T

= 100.93 x 10⁻⁷ T .