Question

Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass through three of the corners of a horizontal square of side 4.0 cm. What is the magnitude of the magnetic field at the fourth (unoccupied) corner of the square due to these wires

Answer 1

B_{total} = (170.71 i ^ - 170.71 j ^ ) 10⁻⁷ T

Step-by-step explanation:

For this exercise, the easiest way to solve it is using Ampere's law, for each wire and then adding vectorly the fields at the point of the free corner

∫B. ds = μ₀ I

Let's use as the surface of calculates a circle of radius at the desired point, the length of this circle is

s = 2π r

substituting

B 2π r = μ₀ I

B =
`(\mu_o \ I)/(2\pi \ r)`

The direction of the field can be found by making the thumb point in the direction of the current and the curved fingers are in the direction of the magnetic field.

Let's find the distance to the unoccupied corner (without wire), using the Pythagorean theorem

for the opposite corner

r₁ =
`√(L^2+L^2)`

r₁ = √2 L

for adjacent corners

r₂ = L

now let's find the magnetic field created for the wire in the opposite corner

B₁ = \frac{\mu_o \ I}{2\pi \ r_1}

let's reduce the distance to the SI system

L = 4.0cm = 0.040m

let's calculate

B₁ = 4π 10⁻⁷ 2.0 /( 2π √2 0.04)

B₁ = 70.71 10⁻⁷ T

the direction of this field is if we suppose that the current is upwards, it is counterclockwise, if the current is downwards, it is clockwise, since all currents go the same direction, all magnetic fields are circular.

Therefore the creed field in the opposite scheme is a perpendicular to the diagonal

The fields produced by the adjacent cables are

B₂ = 4π 10⁻⁷ 2.0 / 2π 0.04

B₂ = 100 10⁻⁷ T

the field is perpendicular to the seam line

the direction of this field is perpendicular to the line that joins the wire with the vacant corner, so we have for each wire

B_{total} = B₂ i ^ - B₂ j ^ + B₁ cos 45 i ^ - B₁ sin 45 j ^

B_{total} = (170.71 i ^ - 170.71 j ^ ) 10⁻⁷ T