Question

Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of the magnetic field halfway between the wires?

Answer 1

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Step-by-step explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

`B = (\mu _oI_1)/(2 \pi r) + (\mu_oI_2)/(2\pi r)`

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

`B = (\mu_o I_1)/(2\pi r) + (\mu_o I_2)/(2\pi r) \\\\B = (\mu_o )/(2\pi r) (I_1 +I_2)\\\\B = (4\pi *10^(-7) )/(2\pi *0.2) (10 +20) = 3.0 *10^(-5)\ T`

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.