Question

The figure below shows two very long straight wires (in cross section) that each carry a current of 4.50 A directly out of the page. Distance d_1 = 8.00 m and distance d_2 = 6.00 m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires? Number units

Answer 1

Here I in both wire = 4.5 A . and Distance between both wire is 8 m so Half of distance= 4 m

Let half point is H. and P from center point H = 6 m.

So direct distance r (between one wire and point P) = 7.22 m (from pythagoras equation whre hyp² =based² +Per²)

Now At P direction of Magnitic field will be perperndicular to r . so it will has two component Bcosθ and Bsinθ .

Due to both wire sin component of B will be in same direction so they will be added.and cos component are in opposite direction so they will be cancled out to each other.

So Net Magnetic Field B = B₁ sin θ + B₂ sin θ

Here current I si same so B₁ = B_{2 ,} So Net B at point P = 2 B sin θ

Now as we know that B=( µ₀ I/ 2πr)

So Net B at P = 2 ( µ₀ I/ 2πr) sin θ = 2 ( ( 4π * 10 ^-7 *4.5 ) / 2 π * 7.22  )* ( 4/7.22) [ because sin θ = per/base]

So B at P = 1.37 * 10 ^-7 T