Question

Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires.

Answer 1

Given:
Magnitude of currents, I₁ = I₂ = 10 A
Separation of the point from the wires, d = 2 cm
Thus, the magnetic field due to current in the wire is given by
$B_1=B_2=\frac{{\mu }_{0}I}{2\mathrm{\pi }d}$


In the figure, dotted circle shows the magnetic field lines due to current carrying wire placed in a plane perpendicular to the plane of the paper.
From the figure, we can see that $∆\mathrm{P}{I}_1{I}_2$ is an equilateral triangle.
$\angle I_1\mathrm{P}{I}_{\mathit{2}}\mathit{}\mathit{=}\mathit{}{60}^{°}$
Angle between the magnetic fields due to current in the wire, θ = $60°$
∴ Required magnetic field at P
$B_{\mathrm{net}}=\sqrt{{B_1}^2+{B_2}^2+2B_1{B}_2\cos \theta }$
$=\sqrt{{\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)}^2+{\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)}^2+\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)+\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)\cos 60°}$
$$\begin{gathered} =\sqrt{\left({10}^{-4}\right)+({10}^{-4}{)}^2+2({10}^{-4}\left)\right({10}^{-4})\times \frac{1}{2}} \\ =\sqrt{3}\times {10}^{-4}\mathrm{T} \\ =1.732\times {10}^{-4}\mathrm{T} \end{gathered}$$