Question

Two long, straight wires, one above the other, are separated by a distance and are parallel to the x-axis. Let the +y-axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current in the +x-direction. What are the magnitude and direction of the net magnetic field of the two wires at a point in the plane of the wires (a) midway between them; (b) at a distance a above the upper wire; (c) at a distance a below the lower wire?

Short answer

a) $B=0$

b)$\overrightarrow{B}=\frac{2{\mu }_{0}I}{3\pi a}\widehat{k}$

c)$\overrightarrow{B}=-\frac{2{\mu }_{0}I}{3\pi a}\widehat{k}$

Step-by-step solution

Magnetic field at the midway

Consider, two long straight wires, one above the other, are separated by a distance 2a and are parallel to the x-axis as shown in the figure. We need to find the magnetic field at the midway between them. The magnetic field due to the upper wire points into the page and due to the lower wire, it points out of the page. Since the two wires carry the same current and in the same direction, then the magnetic field at the midway between them will cancel. So, B = 0 .

Find the magnetic field at a distance a above the upper wire

For a long wire, the magnetic field is given by$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi r}}$ .The magnetic field will be$B=B_1+B_2$ where,

$\begin{array}{r}{B}_1=\frac{{\mu }_{0}I}{2\pi a}\\ B_2=\frac{{\mu }_{0}I}{2\pi \left(3a\right)}\\ B=B_1+B_2\\ B=\frac{{\mu }_{0}I}{2\pi a}\left(1+\frac{1}{3}\right)\\ \overrightarrow{B}=\frac{2{\mu }_{0}I}{3\pi a}\widehat{k}\end{array}$

Positive magnetic field indicates that the field lines are coming out of the page.

Find the magnetic field at a distance a below the lower wire

$\begin{array}{r}B=B_1+B_2\\ B=\frac{{\mu }_{0}I}{2\pi a}+\frac{{\mu }_{0}I}{2\pi \left(3a\right)}\\ \overrightarrow{B}=-\frac{2{\mu }_{0}I}{3\pi a}\widehat{k}\end{array}$

Negative sign indicates that the field lines are going into the page.