A long wire carrying a 5.8 A current perpendicular to the xy-plane intersects the x-axis at x=−2.3cm. A second, parallel wire carrying a 3.0 A current intersects the x-axis at x=+2.3cm. Required:
a. At what point on the x-axis is the magnetic field zero if the two currents are in the same direction?
b. At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions?

a) v    r = 0.7318 cm , b)  r = 7.23 cm

Explanation:

The magnetic field generated by a wire carrying a current can be found with Ampere's law

       ∫ B. ds = μ₀ I

the length of a surface circulates around the wire is

    s = 2π r

where r is the point of interest of the calculation of the magnetic field

         B = μ₀ I / 2π r

In this exercise we have two wires, write the equation of the magnetic field of each one

wire 1     I = 5.8 A

         B₁ = μ₀ 5,8 / 2π r₁

wire 2    I = 3.0 A

         B₂ = μ₀ 3/2π r₂

the direction of the field is given by the rule of the right hand, the thumb indicates the direction of the current and the other fingers the direction of the magnetic field

Let's apply these expressions to our case

a) the two streams go in the same direction

     using the right hand rule for each wire we see that between the two wires the magnetic fields have opposite directions so there is some point where the total value is zero

          B₁ - B₂ = 0

           B₁ = B₂

         μ₀ 5,8 / 2π r₁ = μ₀ 3 / 2π r₂

          5.8 / r₁ = 3 / r₂

          5.8 r₂ = 3r₁

the value of r is measured from each wire, therefore

        r₁ = 2.3 + r

        r₂ = 2.3 -r

we substitute

          5.8 (2.3 - r) = 3 (2.3 + r)

           r (3 + 5.8) = 2.3 (5.8 - 3)

           r = 2.3 2.8 / 8.8

           r = 0.7318 cm

b) the two currents have directional opposite

with the right hand rule in the field you have opposite directions outside the wires

suppose it is zero on the right side where the wire with the lowest current is

         B₁ = B₂

        5.8 / r₁ = 3 / r₂

        5.8 r₂ = 3 r₁

         r₁ = 2.3 + r

         r₂ = r - 2.3

        5.8 (r - 2.3) = 3 (2.3 + r)

        r (5.8 -3) = 2.3 (3 + 5.8)

        r = 2.3 8.8 / 2.8

        r = 7.23 cm

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

d. air pressure decreases as altitude increases.